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## Karnataka 2nd PUC Physics Question Bank Chapter 4 Moving Charges and Magnetism

### 2nd PUC Physics Moving Charges and Magnetism One Marks Questions and Answers

Question 1.

Name the physicist who in his experiment concluded that moving charges or currents produce magnetic fields in the surrounding space.

Answer:

Hans Christian Oersted, (in the year 1820).

Question 2.

What is the source of magnetic field?

Answer:

A charged particle in motion produces a magnetic field around it.

Question 3.

Give the expression for magnetic field at the centre of a circular coil carrying current.

Answer:

Question 4.

At which point around a current carrying circular conductor, the magnetic flux density remains maximum?

Answer:

At the centre of the coil, the magnetic flux will remain maximum. B α \(\frac{1}{x^{3}}\) for x>>>r. Where, ‘r’

is the radius of the coil and ‘x’ is the distance of a point along the axis and away from the centre.

Question 5.

Write the expression for magnetic flux density at one end of a long solenoid carrying current.

Answer:

B = \(\frac{\mu_{0} n i}{2}\) Where, ‘n’ is the linear no. density of the coil winding. n = \(\frac{N}{l}\), N = no. of turns.

Question 6.

Give the expression for magnetic flux density at a point due to an infinitely long conductor perpendicular to it.

Answer:

Question 7.

Give the expression for magnetic flux density at a point at one end of an infinitely long conductor.

Answer:

Question 8.

Write the expression for the magnetic field at a point due to a current clement.

Answer:

Question 9.

Give Biot – Savart formula in vector form.

Answer:

Question 10.

Write the expression for magnetic field at the centre of a current carrying circular coil.

Answer:

Question 11.

Give the expression for magnetic flux density, at a point on the axis, due to a current carrying circular conductor.

Answer:

Question 12.

What is magnetic flux density?

Answer:

The number of magnetic lines passing normally through the surface is known as magnetic flux density.

Question 13.

State Maxwell’s right handed cork screw rule to find the direction of magnetic field around a current carrying conductor.

Answer:

If a right handed cork screw is rotated such that its tip advances in the direction of current, then the magnetic flux density is given in the direction of rotation of the screw head.

Question 14.

Define SI unit of magnetic flux density.

Answer:

If a charge of 1 C is moving at right angles to the direction of magnetic field and experiences a force of 1 N in a direction perpendicular to it, then the applied magnetic flux is said to be 1 tesla or 1 Wbm^{-2}.

Question 15.

What is the action of a magnetic field on a current carrying conductor?

Answer:

A magnetic field exerts a force on a current carrying conductor.

Question 16.

Write the expression for the force on a current carrying conductor placed in a magnetic field.

Answer:

F = BIL sin θ

Question 17.

When is the force on a current carrying conductor (a) maximum (b) minimum?

Answer:

The force on a current carrying conductor is maximum when θ = 90° and minimum when θ = 0° where θ – the angle between conductor and magnetic field.

Question 18.

Which rule is used to determine the force on a current carrying conductor placed in a magnetic field?

Answer:

Fleming’s Left hand rule.

Question 19.

State Fleming’s left hand rule or motor rule.

Answer:

Imagine the first three fingers of left hand stretched at right angles to one another. If the fore finger is pointing in the direction of the magnetic field, and the middle finger is pointing in the direction of the current, then the thumb gives the direction of the force on the conductor.

Question 20.

Write the expression for force between the two long parallel straight conductors carrying current.

Answer:

Question 21.

Define ampere.

Answer:

One ampere is that steady current which when maintained in each of two infinitely long straight parallel conductors of negligible cross section placed 1 m apart in vacuum, produces between the conductors a force of 2 × 10^{-7} newton per metre length of each.

Question 22.

What is the nature of force between two parallel wires carrying current in opposite directions?

Answer:

Two parallel wires carrying current in opposite directions repel and try to move away from each other.

Question 23.

What is the nature of force between two parallel wires carrying current in the same direction? (March 2014)

Answer:

Two parallel wires carrying current in the same direction get attracted towards each other.

Question 24.

What is the principle of an electric motor?

Answer:

A current carrying conductor placed in a magnetic field experiences a force. This is the principle on which an electric motor works.

Question 25.

Name any one use of a moving coil galvanometer.

Answer:

A moving coil galvanometer is used in the measurement of very low currents of the order of 10^{-6} A.

Question 26.

Which instrument can be used for the measurement of current of the order of 10^{-9} A?

Answer:

A suspended type moving coil galvanometer.

Question 27.

What is the torsional couple due to?

Answer:

The torsional couple is due to the elasticity possessed by the phosphor bronze ribbon from which the coil is suspended.

Question 28.

Define current sensitivity of a moving coil galvanometer.

Answer:

Current sensitivity of a moving coil galvanometer is the deflection in mm on a scale placed 1 m from the galvanometer when a current of 1 μA flows through the coil.

Question 29.

How is a galvanometer converted into an ammeter?

Answer:

A galvanometer is converted into an ammeter by connecting a low resistance called shunt in parallel with it.

Question 30.

What is the resistance of an ideal ammeter?

Answer:

The resistance of an ideal ammeter is zero.

Question 31.

Why is an ammeter always connected in series?

Answer:

An ammeter is always connected in series because it has a very low resistance. Introduction of the ammeter in a circuit will not alter the resistance and current.

Question 32.

Why is a voltmeter always connected in parallel?

Answer:

A voltmeter has a very high resistance. It will not alter the resistance and current in the circuit.

Question 33.

What is the resistance of an ideal voltmeter?

Answer:

The resistance of an ideal voltmeter is infinity.

Question 34.

What is a shunt?

Answer:

A shunt is a low resistance, connected in parallel with the galvanometer.

Question 35.

How is a galvanometer converted into a voltmeter?

Answer:

A galvanometer is converted into a voltmeter by connecting a suitable high resistance in series with the galvanometer.

Question 36.

State the principle of superposition of magnetic field.

Answer:

The superposition theorem states that total magnetic field due to several sources, is the vector addition of the magnetic field of each of the individual sources.

Question 37.

Mention the order of magnetic flux density on the surface of a neutron star.

Answer:

The order of magnitude of magnetic flux density on the surface of a neutron star is 10^{8}T.

Question 38.

Represent graphically the directions of \(\vec{v}\), \(\vec{B}\) and \(\vec{F}\) on a moving charge q.

Answer:

Question 39.

Give the expression for the electric force on a moving charge in a uniform electric field.

Answer:

\(\vec{F}\)_{electric} = q \(\vec{E}\) so that F = qB where \(\vec{F}\) and \(\vec{E}\) are magnitudes of F and E respectively.

Question 40.

Give the expression for the magnetic force on a moving charge in an uniform magnetic field. What will be the maximum magnetic force on the moving charge? (July 2014)

Answer:

\(\vec{F}\)_{magnetic} = q \((\vec{v} \times \vec{B})\) S0 that F = qvB sinθ; F_{max} = qvB for θ = 90°.

Question 41.

An electron is accelerated in the direction of the electric field. What will happen to its acceleration over a period of time?

Answer:

It will decrease.

Question 42.

Give the expression for the centripetal force experienced by a moving charge entering into a uniform magnetic field.

Answer:

Centripetal force \(\frac{m v^{2}}{r}\) = qvB sin θ

Question 43.

Mention the direction of the velocity of a charged particle with respect to the magnetic force experienced by it.

Answer:

The direction of \(\vec{v}\) and \(\vec{B}\) are at right angles to each other.

Question 44.

If the magnitude of electric force and magnetic force are equal and if v of a charged particle is perpendicular to both \(\vec{E}\) and \(\vec{B}\), then find the net force on the charge.

Answer:

Zero.

Question 45.

What is meant by the velocity selector?

Answer:

In the presence of crossed fields, only those particles having the ratio v = \(\frac{E}{B}\) pass undeflected. This ratio that serves as a bypass for particles having a constant (E/B) ratio is known as the velocity selector.

Question 46.

What is a cyclotron?

Answer:

A cyclotron is a machine used to accelerate the charged particles or ions to high energies.

Question 47.

Who invented the cyclotron?

Answer:

The cyclotron was invented by E.O. Lawrence and M.S. Livingstone in 1934 to investigate nuclear structure.

Question 48.

Give the expression for cyclotron frequency.

Answer:

v_{c} = \(\frac{\dot{q} \mathbf{B}}{2 \pi m}\)

Question 49.

Give the expression for velocity of a charged particle in a cyclotron.

Answer:

v = \(\frac{q \mathrm{BR}}{m}\) where R is the radius of the trajectory at exit.

Question 50.

Write any one application of the cyclotron.

Answer:

The cyclotron is used to bombard nuclei of atoms of an element and study the resulting nuclear reactions.

Question 51.

Name any one magnetic system.

Answer:

Solenoid and toroid.

Question 52.

What is a toroid?

Answer:

A toroid is a hollow circular ring on which a large number of turns of a wire are closely wound.

Question 53.

Give the expression for magnetic field due to a toroid.

Answer:

B = \(\frac{\mu_{0} \mathrm{NI}}{2 \pi r}\) where r is the radius of a torid.

Question 54.

What is the value of magnetic field at any point in the open space inside a toroid?

Answer:

Zero.

Question 55.

Define the term gyromagnetic ratio.

Answer:

Gyromagnetic ratio is defined as the ratio of the magnetic moment to the anguar momentum

i.e., \(\frac{\mu_{l}}{l}=\frac{e}{2 m_{e}}\) is known as gyrometric ratio of electron.

Question 56.

Mention the value of gyromagnetic ratio for an electron.

Answer:

The gyromagnetic ratio for an electron is 8.8 × 10^{10} C kg^{-1}.

Question 57.

Give the expression for Bohr magneton.

Answer:

Question 58.

Write the value of the magnetic moment of an electron related to orbital motion and spin motion.

Answer:

(µ_{e}) of electron related to orbital motion = 9.27 × 10^{-24} Am^{2}.

Note: µ_{1} of spin motion of electron is also 9.27 × 10^{-24} Am^{2})

Question 59.

Does an electron spin about an axis of rotation?

Answer:

No. It is only an hypothetical situation.

Question 60.

What is meant by spin magnetic moment?

Answer:

Magnetic moment associated with the electron due to spin motion is called as spin magnetic moment.

Question 61.

Define current sensitivity of a moving coil galvanometer.

Answer:

The number of divisions to which a given MCG gets deflected per one ampere of current in it is known as the current sensitivity of MCG.

Question 62.

How does current sensitivity of a MCG depend on couple per unit twist of the wire?

Answer:

Current sensitivity of a MCG is inversely proportional to couple per unit twist of the suspension wire.

Question 63.

How is current sensitivity of a MCG related to the magnetic flux density surrounding the coil?

Answer:

The current sensitivity is directly proportional to the magnetic flux density.

Question 64.

If the current sensitivity gets doubled then what happens to the voltage sensitivity of the MCG?

Answer:

The voltage sensitivity remains the same.

Question 65.

Define voltage sensitivity of a MCG

Answer:

The amount of deflection per unit voltage difference between the terminals of a MCG is known as voltage sensitivity.

Question 66.

Write the expression, for nuclear magnetic moment.

Answer:

Question 67.

Give the expression for figure of merit or current sensitiveness of a galvanometer.

Answer:

The ratio \(\frac{I}{\theta}=\frac{K}{N A B}\) is known as the current sensitiveness.

Question 68.

Give the expression voltage sensitiveness of a galvanometer

Answer:

The ratio \(\frac{V}{\theta}=\left(\frac{K}{N A B}\right) R\) is known as the voltage sensitiveness.

### 2nd PUC Physics Moving Charges and Magnetism Two Marks Questions and Answers

Question 1.

State the Right-hand clasp rule.

Answer:

If a straight conductor carrying current is imagined to be held in the right hand with the thumb pointing the direction of current, then the four fingers clasping the conductor show the direction of the magnetic field.

Question 2.

What is meant by ‘magnetic dipole moment’ of current loop? Write its SI unit.

Answer:

The product of current and area of the loop determines the magnetic dipole moment. Current

in the loop generates a magnetic field. Hence the current loop acts as a magnet. The magnetic moment associated with the current loop is known as magnetic dipole moment.

M = (I) (A) where, ‘A’ – area of the current loop.

M = πr^{2}l where, ‘r’ – is the radius of the circular loop.

Question 3.

Compare the ratio of magnetic field produced at the centre of a circular conductor to a point on the axis of a conductor.

Answer:

where ‘r ’ is the radius of a coil and ‘x’ is the distance of a point from the centre of a coil.

Question 4.

Explain the mechanical effect produced by an electric current.

Answer:

A magnetic field exerts a force on a moving charge. Charges in motion constitute an electric current. Hence a current-carrying conductor placed in a magnetic field experiences a force. If free to move, the conductor moves in the direction of the force.

Question 5.

Give any two differences between an ammeter and a voltmeter.

Answer:

Ammeter:

- It is a device used to measure current in the circuit.
- The resistance of an ideal ammeter is zero.

Voltmeter:

- It is a device used to measure p.d. in a circuit.
- The resistance of an ideal voltmeter is infinity.

Question 6.

What is the nature of the force between the two wires carrying current in

(a) the same direction

(b) opposite directions?

Answer:

(a) The two parallel wires carrying current in the same direction, attract each other.

(b) The two parallel wires carrying current in the opposite directions, repel each other.

Question 7.

Why should the effective resistance of an ideal ammeter be zero and that of a voltmeter, be infinity?

Answer:

The effective resistance of an ideal ammeter should be zero because it should draw the entire current when it is connected in series in the circuit. The resistance of the ideal voltmeter is infinite because it should not draw any current when it is connected in parallel across the circuit component, and hence its introduction should not alter the potential difference to be measured.

Question 8.

Write the expression for the torque on a current loop placed in a uniform magnetic field and explain the terms used.

Answer:

τ = MB sin θ where magnetic moment M = 14.

where, I – current, A – area of loop = πr^{2}.

τ_{max} = MB, τ_{min} = 0

Question 9.

Write the expression for the magnetic field at a point on the axis of a long solenoid carrying current and give the meaning of the symbols used.

Answer:

B = μ_{0}ni, Here B ∝ n and B ∝ i. n – number density of turns, i – current.

Question 10.

Give the expression for Lorentz force acting on a moving electric charge in a combined electric and magnetic field.

Answer:

The combined electric and magnetic forces acting on a moving electric charge is called the Lorentz force.

Question 11.

An electron in an electric field is made static and prevented from falling under gravity. Write the equation satisfying the above condition.

Answer:

Question 12.

What is the cause for the helical motion of a charged particle in a magnetic field?

Answer:

The components of velocity, perpendicular to the magnetic field cause a circular motion of the charged particle and if there exists a component of velocity along the direction of the magnetic field, then the charged particle describes a helical path.

Question 13.

What is meant by the pitch of a charged particle describing a helical path?

Answer:

The distance moved along the magnetic field in one rotation is called the pitch p. i.e., p = v_{h} T = \(\frac{2 \pi m v_{h}}{q B}\)where ‘v_{A}’ is the horizontal component of the velocity of the charged

particle.

Question 14.

Give the direction of current passing through an open surface.

Answer:

If the fingers of the right hand be curled in the sense the boundary is traversed in the loop integral, then the direction of the thumb gives the sense in which the current is regarded as positive.

Question 15.

State Ampere’s circuital law and give the mathematical expression for the circuital law.

Answer:

The integral of the products of tangential component of magnetic field and the element of length in a closed loop is equal to μ_{0} times the total current passing through the surface.

Question 16.

Write the equation for the force between the two parallel straight wires and give the meaning of the symbols used.

Answer:

F = \(\frac{\mu_{0} I_{a} I_{b}}{2 \pi r} L\) where I_{a}, I_{b} are the respective currents in the wires, r is the distance between the two straight parallel conductors and Lis the length of the wires.

Question 17.

Give the direction of the magnetic field due to a current carrying circular conductor.

Answer:

The direction of magnetic field will be in a plane at right angles to the plane of the coil and points outward away from the centre for an anticlockwise current and inward towards the centre for a clockwise current.

Question 18.

Can a current carrying circular loop act as a magnetic dipole? Give the expression for the magnetic field at a point, due to a current carrying circular loop.

Answer:

Yes, B = \(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 m}{x^{3}}\)

where, m = magnetic moment of the current loop

m = IA

I = current,

A = area of circular loop,

x = distance of the point on the axis of the coil.

Question 19.

Give an analogy for the torques experienced by an electric dipole and a magnetic dipole.

Answer:

Question 20.

Write the expression for the magnetic moment due to an electron circulating around the nucleus of an atom.

Answer:

Magnetic moment = Iπr^{2} = \(\frac{e v R}{2}\) The direction of this magnetic moment is into the sheet of paper for clockwise current (due to anticlockwise motion of electron).

Question 21.

Represent the magnetic moment vector in terms of angular momentum vector of electrons revolving around the nucleus.

Answer:

where -ve sign indicates that the direction of angular momentum is opposite to the magnetic moment.

### 2nd PUC Physics Moving Charges and Magnetism Three Marks Questions and Answers

Question 1.

If ‘P’ is the middle point on the axis of a solenoid then derive the expression for magnetic field at that point.

Answer:

Question 2.

What are the factors which determine the force on a current carrying conductor in a magnetic field?

Answer:

The factors affecting the force are

- length of the conductor.
- current and
- magnetic field.

Question 3.

A galvanometer has a resistance G and requires a current ‘g for full scale deflection. How do you convert it into

(i) an ammeter?

(ii) a voltmeter?

Answer:

(a) S = \(\frac{I_{g} G}{I-I_{g}}\) i.e., I ∝ g where ‘S’ is the shunt resistance connected in parallel to the galvanorneter and

(b) R = \(\frac{V}{I_{g}}\) – G i.e., I_{g} ∝ V where ‘R’ is a resistance connected in series with the galvanometer.

Question 4.

Describe an experiment conducted by Oersted to demonstrate that magnetic field is produced by electric current.

Answer:

Consider a wire PQ connected to a cell with a switch. Let a magnetic needle be placed below the needle. If the switch is opened, no defletion was observed. However if the switch is closed, then north pole of the magnetic is deflected whose direction is given by the right hand clasp rule.

If the magnitude of emf is increased, the deflection also increases. On reversing the polarity of the emf, the direction of deflection is also reversed. This experiment clearly demonstrates

the magnetic effect of electric current.

### 2nd PUC Physics Moving Charges and Magnetism Five Marks Questions and Answers

Question 1.

Obtain an expression for the magnetic force on a current carrying conductor.

Answer:

Let l be the length of a conductor. Let A be its area of cross section. Let n be the number density of mobile charge carriers in the conductor.

The total number of mobile charge carriers is nAl.

Let the average drift velocity be ‘v’_{d} corresponding to the current I.

Let B be an external magnetic field surrounding the conductor.

Magnetic force on these carriers is,

By definition current density j = \(\frac{1}{A}\)

and q = e (charge on the electron)

The direction of current is in the direction of \(\vec{l}\) vector, \(\vec{l}\) = ĵ and jA = I = current.

Question 2.

Obtain expressions for the radius of the circular path and its frequency, of a charged particle entering into an uniform magnetic field.

Answer:

Let B represent the uniform magnetic field applied perpendicular to the plane of the paper and pointing inwards.

Let q represent a +ve charge entering the field with a velocity v at right angles to the field. Let F represent the magnetic force on the charged particle, which acts perpendicular to the direction of velocity. No work is done by the force.

Magnetic force \(\overrightarrow{\mathrm{F}}=q \vec{v} \times \overrightarrow{\mathrm{B}}\). This force acts as a centripetal force and produces a circular motion (as per the Fleming’s Left Hand Rule).

Therefore, radius of the circular path r ∝\(\frac{m}{q}\) for a given uniform speed and magnetic field.

Period of revolution T = \(\frac{2 \pi r}{v}=\frac{2 \pi m}{q \mathrm{B}}\) and frequency v = \(\frac{1}{T}=\frac{q B}{2 \pi m}\)

In the presence of the component of velocity along B, the path described by the charged particle will be helix and whose pitch p = v_{h}T; v_{h} is the horizontal component of velocity along B.

Question 3.

Describe the motion of a charged particle ¡n a combined electric and magnetic field.

Answer:

Let E be the uniform electric field applied in the y direction and B be the uniform magnetic field applied along the z direction.

Let th charge q move with a speed y along the x direction.

By applying Fleming’s Left Hand Rule, we note that the resultant force on th charged particle

If qE = qvB, then net force on the charge will be zero and the charge continues to move along x direction without deflection.

Since qE = qvB, v = \(\frac{E}{B}\) where V is called the velocity. E and B are called crossed fields.

The crossed fields \(\vec{E}\) and \(\vec{B}\) may be used to select charged particles of a particular velocity out of a beam containing charges moving with different speeds.

Question 4.

Explain briefly the action of cyclotron.

Answer:

- Cyclotron is a device to accelerate charged particles or ions to high energy levels. Cyclotron uses both electric and magnetic fields to increase the energy of the charged particles.
- The frequency of motion of charged particles is independent of the radius of the circular path and energy.
- Cyclotron uses an oscillator and its voltage is adjusted so that the polarity of the ‘dees’ is reversed during the period in which the particle completes one half cycle
- Cyclotron consists of 2 ‘D’ shaped semicircular disc like metal containers D, and Dr
- Magnetic field is applied such that it is perpendicular to the paper and points out of the paper D
_{1}and D_{2}are connected to the oscillator. The deflection plate makes the charged particles to travel straight and exit out of the port with greater speed and energy.

The phase of the oscillator is so adjusted (Φ_{applied}= Φ_{cyclotron frequency}) that the charged particles move from D_{1}to D_{2}. The charged particles spiral until they acquire a high energy level before leaving the cyclotron.

Cyclotron frequency, V_{c}= \(\frac{q \mathrm{B}}{2 \pi \mathrm{m}}\)

but velocity of charged particle,

v = \(\frac{q \mathrm{BR}}{m}\) where R, is the radius of a dee. Hence K.E. of charged particle = \(\frac{1}{2}\)mv^{2}

Question 5.

Give any five practical applications of cyclotron or highly energetic charged particles obtained from a cyclotron. (March 2014, July 2014)

Answer:

- Cyclotron is used to bombard nuclei with energetic particles.
- Cyclotron is used to implantions into solids.
- Cyclotron is used to modify the properties ofa material by bombarding it with the accelerated charged particles.
- Cyclotron is used to snthçsize new materials.
- Cyclotron is used in hospitals to produce radioactive substances.

Question 6.

State and explain Biot-Savart’s Law.

Answer:

Biot – Savart’s law (Laplace rule) states that the magnetic flux density (dB) at a point due to a small current element (idl) of a current carrying conductor is directly proportional to

(i) the current

(ii) the length of the element ‘dl’,

(iii) sine of the angle made between the current element and the line joining the point and is inversely proportional to

(iv) the square of the distance between the point and the element.

Question 7.

Obtain an expression for the magnetic field on the axis of a circular current loop. (March 2014)

Answer:

Let R be the radius of a current loop, carrying current I. Let R be a point on the axis of a conductor. Let dB be magnetic field at P, due to a current element ‘idl’.

From the figure, θ + α = 90°, so that α = 90° – θ

and cosα = cos (90°- θ) = Sin θ = \(\frac{R^{2}}{\left(R^{2}+x^{2}\right)^{\frac{1}{2}}}\) …. (1)

Let dB be the horizontal component of dB.

Applying Biot-Savart’s law,

Question 8.

Obtain an expression for the magnetic dipole moment of a revolving electron in a hydrogen atom, and hence find the value of Bohr magneton.

Answer:

Let e represent the charge on the particle electron.

Let r be the radius of a circular path. Let the electron revolve in an anticlockwise direction. Conventional current is taken as clockwise. So magnetic moment can be visualised to be pointing inwards towards the centre and perpendicular to the plane of the orbit.

By definition, electric current

i.e., I = \(\frac{e}{\mathrm{T}}\) where, T is the time period of revolution of electron

The period of revolution of electron,

Magnetic moment associated with the orbiting electron,

m = IA = µ_{1}

where, µ_{1} = magnetic moment due to orbital motion.

Hence, µ_{1} = \(\left(\frac{e v}{2 \pi r}\right)\left(\pi r^{2}\right)\)

i.e., µ_{1} = \(\left(\frac{e}{2 m_{e}}\right)\left(m_{e} v r\right)\) where, angular momentum l = m_{e}vr

∴ µ_{1} = \(\left(\frac{e}{2 m_{e}}\right) l\)

The direction of \(\vec{\mu}_{l}\) is opposite to the direction of angular momentum.

The expression µ_{1} = \(\frac{e v r}{2}\) for I orbit electron is called Bohr magneton.

µ_{1} = 1.6 × 10^{-19} × 2.18 × 10^{6} × 0.53 × 1o^{-10} ÷ 2 = 9.24 × 10^{-24}Am^{2}.

Question 9.

Obtain an expression for magnetic field in terms of magnetic dipole moment associated with circular current loop.

or

Show that circular current loop can be associated with a magnetic dipole.

Answer:

We know that the magnetic field at a point on the axis of a circular loop.

Hence circular loop acts as a magnetic dipole.

Note :

As an analogous to the electric field intensity

Comparing (1) and (2) we note that by substituting µ_{0} as \(\frac{1}{\varepsilon_{0}}\)

and \(\vec{B}\) by \(\vec{E}\), \(\vec{m}\) by \(\vec{p}_{e}\), one can relate the magnetic and electric fields due to the corresponding dipoles.

Question 10.

Obtain an expression for the magnetic field induction at a point due to an infinitely long thin conductor or wire by applying Ampere’s circuital law.

Answer:

Let I be the current in the wire PQ. Let P be a point at a perpendicular distance r from the wire. Let dl be the element of length of the closed path. The direction of the magnetic field will be tangential to the point P on the elements

Applying Ampere’s circuitaldaw, we have

Note : B along the axis of the wire = 0

Question 11.

Obtain an expression for Magnetic field inside a solenoid by using Ampere’s Circuital Law.

Answer:

Consider a rectangular Amperian loop abcd. Along cd the field is zero because the field at the exterior is weak and moreover the field is along the axis of the solenoid with no perpendicular or normal component. Therefore the field outside the solenoid approaches zero.

Along transverse sections bc and ad, the field component is zero.

Let the field along ab be B. The relevant length of the Amperian loop is l.

Let n be the number of turns per unit length.

The total number of turns corresponding to the length l is nl. The enclosed current I_{e} = Inl where I is current in the Solenoid.

From Ampere’s circuital law B∫dl = µ_{0}I_{e}

i.e., BI = µ_{0}Inl. Therefore B = µ_{0}nI.

The direction of magnetic field is given by the right-hand rule.

Note: The solenoid is used to obtain a uniform magnetic field.

Question 12.

Obtain an expression for magnetic field inside the toroid.

Answer:

Consider three circular Amperian loops 1,2 & 3 as shown in the figure by dotted line.

Let the magnetic field along loop 1 be B_{1} in magnitude. The length of the loop ∫dl = 2πr_{1}. This current loop does not enclose current.

Therefore I_{e} = 0 and B_{1} = 0. Therefore, the magnetic field at any point P in the open space inside the toroid is zero. Consider the Amperian loop 3. Let B_{3} be the magnetic field along the loop.

Here ∫dl = 2πr_{3} ; The current coming out of the planes is cancelled exactly by the current going into it. Therefore I_{e} = 0 and B_{3} at any point Q outside the toroid is zero.

Here ∫dl = 2πr_{3} The current coming o3 at any point Q outside the toroid is zero.

Let the magnetic field inside the toroid beB. The length of Amperian in loop 2 = ∫dl = 2πr_{2}.

The current enclosed for N turns of toroidal coil is NI.

From Amperian Circuital law,

B∫dl = µ_{0}I_{enclosed}

i.e., B(2πr_{2}) = µ_{0}NI

i.e., B = \(\frac{\mu_{0} N I}{2 \pi r_{2}}\)

Let the number of turns per unit length be n. Then \(\frac{N}{2 \pi r_{2}}\) = n.

Thus, B = µ_{0}NI.

Question 13.

Derive the expression for the force between two parallel straight conductors carrying current and hence define‘ampere’.

Answer:

X and Y are two long straight parallel conductors carrying currents I_{1} and I_{2} irrespectively, and placed close to each other. d is the separation between the two conductors and L is the length of the conductors.

The magnetic field at any point on the conductor Y due to current I_{1} in the conductors X, is given by B_{1} = \(\frac{\mu_{0}}{4 \pi} \frac{2 I_{1}}{d} \cdot \vec{B}_{1}\) acts in a direction perpendicular to the plane containing the two conductors. The conductor Y which carries current I_{2}, experiences a mechanical force due to B_{1} acting on it and this force is given by F_{1} = B_{1} I_{2}, L sin θ.

According to Fleming’s left hand rule, the direction of the force F_{1} on X is perpendicular’to B_{2} and is towards the conductor Y.

The magnetic field at any point on the conductor X due to the current I_{2} in y, is given by

B_{2} = \(\frac{\mu_{0}}{4 \pi} \frac{2 I_{2}}{d} \cdot \vec{B}_{2}\) acts on X and opposite to B1. The mechanical force acts on X due to B_{2} is I_{1} L sin θ.

According to Fleming’s left hand side rule, the direction of the force F_{2} on X is perpendicular to X and it is towards the conductor y if the current I_{1} is inwards (or away from the conductor Y if the current I_{1} is outwards)

The Force F_{1} acting on a certain length of the conductor Y due to the current in the conductor X is equal in magnitude to the force F_{2} acting on the same length of X due to the current in conductor Y. If the two conductors carry the currents in the same direction (parallel currents) then the forces attract each other. If the two conductors carry the currents in the opposite directions (anti parallel currents), then they are found to repel each other because the two forces act away from each other.

The force per unit length on each conductor is

One ampere of current can be defined as that constant current which when maintained through each of the two infinitely long straight parallel conductors of negligible area of cross section in the same direction placed 1 metre apart in vacuum, causes an attractive force of 2 × 10^{-7} N m^{-1} length on each conductor.

Question 14.

Obtain an expression for torque acting on a rectangular current loop.

Answer:

Let a and b represent breadth and length of a rectangular loop.

Let I be the current in the coil. Let m be the magnetic moment, τ be the torque on the coil and B be the magnetic flux linked with the coil. Let B and s represent brush and slip ring.

By applying Fleming’s Left Hand Rule, we note that force on AB and CD remains the same but act opposite to each otfter. These two and opposite forces constitute couple.

Deflecting couple or torque is given by the product of magnitude of any one force and perpendicular distance between the forces.

Torque, τ = Fa

where, Blbsinθ = F on AB or CD.

= (Blb)a sinθ = IB(ab)sinθ.

i.e.. τ = IBAsinθ

where area of the loop = A = ab

If θ = 0, sinθ = 0 then torque τ = 0

But magnetic moment of the current loop ABCD = IA = m

Hence torque on the current loop = τ = mBsinθ

\(\vec{\tau}=\vec{m} \times \vec{B}\)

Note : As an analogue to torque on electric dipole

\(\vec{\tau}=\vec{p}_{e} \times \overrightarrow{\mathrm{E}}\)

where, p_{e} -dielectric moment, \(\vec{E}\) = Electric field

Question 15.

With the help of a neat labelled diagram explain MCG and obtain an expression for current sensitivity of MCG

Answer:

NS – poles of powerful horseshoe magnet

FMS – Ferromagnetic substance

S – annular spring

P- pointer

JB-jewel bearing

C – coil

S – linear scale

FL – fieldlines

We know that torque acting on the coil τ = NIAB – (This is also known as the deflecting couple)

Restoring couple is due to the spring with couple per unit twist k.

If Φ is the deflection in radian then at equilibrium

Restoring couple = deflecting couple

KΦ = NIAB

Hence θ = \(\frac{\mathrm{NIAB}}{\mathrm{K}}\)

The ratio \(\frac{\theta}{I}=\frac{N A B}{K}\) is known as the current sensitivity. If R is the electrical resistance of the moving coil then V = IR or I = \(\frac{V}{R}\)

Question 16.

Explain how a galvanometer is converted into an ammeter.

Answer:

To convert a galvanometer into an ammeter, a low resistance called a shunt resistance is connected in parallel with the galvanometer. A galvanometer with this modification is called an ammeter.

The value of the shunt resistance S depends on the maximum current to be measured.

If I be the maximum current to be measured, I_{g} be the current for a full scale deflection and G the resistance of the galvanometer then,

from equation (1) I_{g} ∝ I

Since S and G are constants, the scale can be graduated to give the main current directly.

Question 17.

Explain how a galvanometer is converted into a voltmeter.

Answer:

To convert a galvanometer into a voltmeter, a high resistance is connected in series with the coil. The galvanometer with this modification is called a voltmeter. The value of the high resistance R to be connected in series with the coil depends on the maximum potential difference to be measured. If Ig be the current for a full scale deflection then,

V = I_{g} (G + R)

G is the resistance of the galvanometer.

∴ R = \(\frac{V}{I_{g}}\) – G

I_{g} ∝ V

Since G and R are constants, the scale can be graduated to read potential differences directly.

A voltmeter js used for measurement of potential difference. It should be connected in parallel in a circuit.

Question 18.

Distinguish between an ammeter and a voltmeter.

Answer:

Ammeter:

- An ammeter is an instrument used to measure the current in a circuit.
- It is always connected in series in the circuit.
- The resistance of the ammeter is very low.
- The resistance of an ideal ammeter is zero.
- The shunt resistor takes the maximum current.

Voltmeter:

- A voltmeter is an instrument used to measure the potential difference across the circuit.
- It is always connected in parallel across the load in the circuit.
- The resistance of the voltmeter is very high.
- The resistance of an ideal voltmeter is infinity.
- The ballast or series resistor takes the maximum voltage.

### 2nd PUC Physics Moving Charges and Magnetism List of Formulae

### 2nd PUC Physics Moving Charges and Magnetism Numericals and Solutions

Question 1.

Two circular coils of mean radii 0.1 m and 0.05 m consisting of 5 turns and 10 turns respectively are arranged concentric to one another with their planes at right angles to each other. If a current of 2A is passed through each of them, calculate the magnitude of the resultant magnetic field at their common centre.

Answer:

Given : r_{1} = 0.10 m; r_{2} = 0.05m, n_{1} = 5; n_{2} = 10; I = 2A

We know that, B at the centre of the circular conductor carrying current

α = tan^{-1} (4) w:r.t. B_{1}

i.e., q = 75°58′ w.r.t. B_{1}

The direction of the resultant is 75°58′ w.r.t. the direction of B_{1}

Question 2.

Find the magnitude of magnetic induction at a point 0.06m from the centre and along the axis of a circular coil carrying a current of 2 A. Also calculate the magnitude of magnetic induction at the centre of the coil.

Given: Number of turns in the coil = 20

Mean radius of the coil = 0.05 m.

Answer:

Given I = 2A, r = 5 × 10^{-2}m, n = 20, d = 6 × 10^{-2}

We know that the magnetic field at the centre of the coil,

Question 3.

A current of 5 mA passing through a coil of 10 turns produces a magnetic f of 6.28 x 1028 T at the centre of the coil. Calculate the radius of the coil.

Answer:

Given i = 5 × 10^{-3} A, n = 10, B = 6.28 × 10^{28} T.

Question 4.

The magnetic flux at two points on the axis of a circular coil at distances 5 cm and 20 cm from the centre are in the ratio 8:1. Find the diameter of the coil.

Answer:

Given x_{1} = 5 cm, x_{2} = 20 cm

Taking cube root on both sides we write

Squaring both sides we get,

3r^{2} = (20)^{2} – 4(5)^{2}

3r^{2} = 300

∴ r = ± 10 cm

Diameter of the T.G. coil = 0.2 m.

Question 5.

A circular coil of 10 turns and mean radius of 0.1 m is kept with its plane in the magnetic meridian. If a current of 2 A passes through it, calculate the resultant magnetic field at its centre. (B_{H} = 4 × 10^{-5} T)

Answer:

Magnetic field at the centre,

Resultant magnetic field = 13.2 × 10^{-5} T.

Question 6.

The plane of a circular coils at right angles to the magnetic meridian. If the number of turns in the coil equals 2, radius of the coil equals 0.078 m and current 2.8 A, then calculate the net magnetic field at the centre for clockwise and anticlockwise currents.

Answer:

Given : B_{H} = 4 × 10^{-5} T.

B_{C} = 4.15 × 10^{-5} T

Case (i): For anti clockwise current

B_{R} = B + B_{H}

BR = (4.51 + 4) × 10^{-5}T = 8.51X10 T

Case (ii): For clockwise current

B_{R} = B – B_{H}

= (4.51 -4) × 10^{-5}

= 0.51 × 10^{-5} T

Question 7.

Two identical circular coils are separated by a distance twice the radius of either of the coil. If n = 2, r = 0.08 m, i = 3A, then calculate the resultant magnetic field at the mid point of the line joining their centres, for currents in the same sense and opposite sense in the two circular coils.

Answer:

Given: n = 2, r = 0.08 , i = 3A

Magnetic field at a point on the axis,

Question 8.

A straight horizontal rod of mass 30 grams and length 0.3m is placed in a uniformhorizontal magnetic field of strength 0.2 T perpendicular to the rod. Calculate the current through the rod, if (he force on it just balances its weight, g = 9.8 ms^{-2}

Answer:

Thc force acting on the rod is given by F=BIL. This force is balances the weight of the rod

BIL = mg

Question 9.

Calculate the scparation between two long, straight parallel wires, carrying currentsm of 100 A and 50 A, if they repel each other with a force of 1 Nm-’.

Answer:

Given I_{1} = 1oo A, I_{2} = 50 A. F = 1 Nm^{-1}

Let d be the separation between the two conductors then,

Question 10.

A conductor of length 10 cm carrying 2.5 A current is placed in a magnetic field of 0.5 T. Calculate the maximum and minimum forces on the conductor. What is the force on the conductor when it is placed at 45° to the direction of field?

Answer:

l = 10 cm = 0.1 m, 1= 2.5 A, B = 0.5 T

F = BIL sin θ

F_{max} = BIL (when θ = 90°)

For θ = 45° F = 0.5 × 2.5 × 0.1 sin 45° = 8.83 × 10^{-2}N.

Question 11.

Calculate the force experienced by the wire (iii) from the figure.

Answer:

Force on (iii) due to (i) = \(\frac{10^{-7} \times 2 \times 15 \times 10}{0.03}\) Nm^{-1} towards (1)

= 1 × 10^{-3}Nm^{-1}

Force on (iii) due to (ii) = \(\frac{10^{-7} \times 2 \times 5 \times 10}{0.02}\) Nm^{-1} away from (ii)

= 0.5 × 10^{-3} Nm^{-1}

Net force on (iii) = (1 – 0.5) × 10^{-3} = 0.5 × 10^{-3} Nm^{-1} towards (i).

Question 12.

A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10T normal to the plane of the coil. If the current in the coil is 5.0 A, then what is the

(a) total torque on the coils

(b) total force on the coil

(c) average force on each electron in the coil due to the magnetic field.

(The coil is made of copper wire of cross-sectional area 10^{-5} m^{2}, and free electron density in copper is 10^{29} m^{-3}.)

Answer:

N = 20, R = 0.10m, B = 0.10 T, I = 5, A = 10^{-5} m^{2}, n = 10^{29} m^{-3}.

Torque = BAN I sinθ.

(a) Torque on the coil = 0, since there will be no rotatory effect of the coil.

(b) Total force on the coil = 0 because the field will be uniform.

(c) magnetic force on each electron F = Bev = \(\frac{B I}{n A}\)

i.e., F = \(\frac{0.10 \times 5}{10^{29} \times 10^{-5}}\) = 5 × 10^{-25} N.

Question 13.

A solenoid 6Q cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near the centre) normal to its axis, both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery, which supplies a current of 6.0 A in the wire. What value of current in the windings of the solenoid can support the weight of the coil? g = 9.8 ms^{-2}.

Answer:

Given : N = 300 × 3 = 900 turns

l = 0.60 m, m = 2.5 × 10^{-3} kg.

∴ n = \(\frac{\mathrm{N}}{l}=\frac{900}{0.6}\)

Magnetic field at the centre of a solenoid = μ_{0} nl

Force experienced by the wire = BI’l’ sin90°

i.e., F = (μ_{0} nl) I’l’

but F’ = mg, where m – mass of the wire

∵ F = F’

μ_{0} nI I’l’ = mg

I = 108 A

Hence a current of 108 A can support the weight of the wire.

Question 14.

A galvanometer coil has a resistance of 12 Ω and the meter shows FSD for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 – 18V.

Answer:

Given G = 12Ω, I_{g} = 3 × 10^{-3} A, V = 18V

Using the formula

Question 15.

A galvanometer coil has a resistance of 15 Ω and the meter shows FSD for a current of 4mA. How will you convert the meter into an ammeter of range 0 to 6A.

Answer:

Given : G = 15Ω, I = 4 × 10^{-3}A, I = 6A

A shunt resistance 0.010Ω is connected in parallel with MCG to convert it into an ammeter of range 0 to 6A.

Question 16.

A toroid has a non ferromagnetic core of inner radius 25 cm and outer radius 26 cm around which 3500 turns of wire are wound. If the current in the wire is 11A, what is the magnetic field

(a) outside the toroid?

(b) inside the core of the toroid?

(c) in the empty space surrounded by the toroid?

Answer:

Mean radius of the toroid = \(\frac{0.25+0.26}{2}\) = 0.25m

i.e., n = 2184. 196

(a) Magnetic field outside the toroid is zero.

(b) Magnetic field inside the core of the toroid,

B = µ_{0}nl = 4 × 3.142 × 10^{-7} × 2184.196 × 11 T .

= 3.02 × 10^{-2} T.

(c) In the empty space surrounded by the toroid, magnetic field = 0

Question 17.

A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires,

(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

(b) What will be the total tension in the wires if the direction of current is reversed, keeping the magnetic field same as before?

Answer:

(a) For zero tension in the wire,

F = mg, m = 60 × 10^{-3}kg.

i.e., BIL sin90° = mg; g = 10 ms^{-2}

or \(\frac{m g}{I L}=\frac{6 \times 10^{-2} \times 10}{5 \times 0.45} \mathrm{T}\)

i.e., B = 0.27 T

(b) If the direction of current is reversed, net force acting downwards = BIL + mg

= mg + mg = 2mg

= 2 × 0.06 × 10

= 0.12 × 10

= 1.2N (Tension (T_{1} + T_{2}) in the wires

= 1.2N.

Question 18.

In a chamber, a uniform magnetic field of 8.0 G is maintained. An electron with a speed of 4.0 × 10^{-6} ms^{-1} enters the chamber in a direction normal to the field.

(a) Describe the path of the electron.

(b) What is the frequency of revolution of the electron?

(c) What happens to the path of electron if it progressively loses its energy due to collisions with the atoms or molecules of the environment?

Answer:

Given : B = 8 × 10^{-4}, v = 4 × 10^{-6} ms^{-1}.

0 = 90°, m_{e} = 9.1 × 10^{-31} kg.

(a) The path of the electron is circular of radius r = \(\frac{m v}{e B}\)

r = 2.84 × 10^{-2}m.

(b) T = \(\frac{2 \pi r}{v}\)

i.e., T = 4.46 × 10^{-2-6} s

T = 4.46 × 10^{-8} s

i.e., Frequency f = 2.24 × 10^{7} Hz.

The frequency is independent of the size of the orbit or energy.

(c) The electrons will follow either a helical or a spiral path.

Question 19.

A square coil of side 10 cm consists of 20 turns and carries a current of 12A. The coil – is suspended vertically and normal to the plane of the coil, makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80T. What is the magnitude of torque experienced by the coil?

Answer:

Area of the square coil = A = l^{2}

A = (0.1m)^{2} = 10^{-2} m^{2}.

N = 20,1 = 12A, θ = 30°, B = 0.80 T; sin30° = 1/2

Torque experienced by the coil,

τ = BANIsinθ.

i.e., τ = 0.8 × 10^{-2} × 20 × 12 × 1/2 Nm

i.e.,τ = 0.96 Nm

Question 20.

Two concentric circular coils X and Y of radii 16 cm and 10 cm respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16A; coil Y has 25 turns and carries a current of 18A. The direction of the current in X is anticlockwise and is clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their Centre.

Answer:

Given : N_{1} = 20, I_{1} = 16 A, R = 0.16 m

B_{x} = 1.257 × 10^{-3} T pointing east.

N_{2} = 25, I_{2} = 18A, R = 0.10 m,

B_{y} = \(\frac{10^{-7} \times 2 \times 3.142 \times 25 \times 18}{0.10}\)

B_{y} = 2.828 × 10^{-3} T pointing west.

Question 21.

An electron emitted by a heated cathode and accelerated through a p.d. of 2.0 kV . enters a region with uniform magnetic field of 0.15T. Determine the trajectory of the electron if the field

(a) is transverse to its initial velocity

(b) makes an angle of 30° with the initial velocity.

Solution,

(a) V = 2 × 10^{-3} V, θ = 90°, B = 0.15T

\(\frac{e}{m}\) = 1.76 × 10^{11} Ckg^{-1} m

The path of the electron will be circular and of radius r = \(\frac{m v}{e \mathrm{B}}\)

(b) If the field makes an angle of 30°, then v_{hor} = vcosθ = 2.64 × 10^{7} × cos30°

i.e., v_{H} = 2.64 × 0.866 × 10^{7} = 2.286 × 10^{7} ms^{-1}

in the direction of B and the path will be helical.

Question 22.

An a – particle of mass 6.65 × 10^{-27} kg is travelling at right angles to a magnetic field with a speed of 6 × 10^{5} ms^{-1}. The strength of the magnetic field is 0.2 T. Calculate the force on the α – particle and its acceleration.

Answer:

Force on a charged particle,

F = Bqv sin θ

Here m = 6.65 × 10^{-27} kg, q_{α} = 2e = 2 × 1.6 × 10^{-19} C

q_{α} = 3.2 × 10^{-19} C

B = 0.2 T, v = 6 × 10^{5} ms^{-1}

F = 0.2 × 3.2 × 10^{-19} × 6 × 105 × sin 90°

= 3.84 × 10^{-14} N

a = 5.77 × 10^{12} ms^{-2}

Question 23.

A beam of protons moving with a velocity of 4 × 10^{5} ms^{-1}, enters a uniform field of 0.3T at an angle of 60° to the direction of the magnetic field. Find

(i) the radius of the helical path of the proton beam and

(ii) pitch of the helix. (Mass of proton = 1.67 × 10^{-27} kg; charge on proton = 1.6 × 10^{-19} C).

Answer: