# NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

Q. A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in midair by a uniform horizontal magnetic field B
(Fig.). What is the magnitude of the magnetic field ?

Ans. In the given figure, there is an upward force F, of magnitude IlB. For mid-air suspension, this must be balanced by the force due to gravity : $$\text{mg}=\text{I/B}\\\text{B}=\frac{mg}{\text{I}l}\\=\frac{0.2×9.8}{2×1.5}=0.65\text{T}$$

• Q. (i) A current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e., turns about the vertical axis)?
• (ii) A current-carrying circular loop is located in a uniform external magnetic field. If the loop is free to turn, what is its orientation of stable equilibrium ? Show that in this orientation, the flux of the total field (external field + field produced by the loop) is maximum.
• (iii) A loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible, why does it change to a circular shape ?
• Ans. (i) No, because that would require τ to be in the vertical direction. But τ = IA × B, and since area A of the horizontal loop is in the vertical direction, τ would be in the plane of the loop for any B.
• (ii) Orientation of stable equilibrium is one where the area vector A of the loop is in the direction of external magnetic field. In this orientation, the magnetic field produced by the loop is in the same direction as external field, both normal to the plane of the loop, thus giving rise to maximum flux of the total field.
• (iii) It assumes circular shape with its plane normal to the field to maximize flux, since for a given perimeter, a circle encloses greater area than any other shape.

Q. A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid ?

Ans. The number of turns per unit length is,

$$\text{n}=\frac{500}{0.5}=1000\space\text{turns/m}\\\text{The length l = 0.5 m and radius r = 0.01 m. Thus l/a = 50 i.e., l \text{\textgreater}\text{\textgreater} a.}\\\text{Magnetic field inside the solenoid is given by}\\\text{B}=\mu_0nI\\=4\pi×10^{-7}×10^{3}×5\\=6.28×10^{-3}\space\text{T}$$

Q. A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire ?

Ans. Given, Length of the wire, l = 3 cm = 0.03 m

• Current flowing in the wire, I = 10 A
• Magnetic field, B = 0.27 T
• Angle between the current and magnetic field, θ = 90°
• Magnetic force exerted on the wire is given as
• F = BIl sin q
• = 0.27 × 10 × 0.03 sin 90°
• = 8.1 × 10– 2 N
• Hence, the magnetic force on the wire is 8.1 × 10– 2 N. The direction of the force can be obtained from Fleming’s left hand rule.

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

• Ans. Given, Length of a side of the square coil, I = 10 cm = 0.1 m
• Current flowing in the coil, I = 12 A
• Number of turns on the coil, n = 20
• Magnitude of the magnetic torque experienced by the coil in the magnetic field
• τ = n BIA sin θ
• Here, τ = torque, I = current
• A = Area of the square coil
• = l × l = 0.1 × 0.1 = 0.01 m2
• So, τ = 20 × 0.8 × 12 × 0.01 × sin 30°
• = 0.96 N m

Q. A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on  different cases shown in given figures ?

• (i) What is the torque on each case ?
• (ii) Which case corresponds to stable equilibrium ?  • Ans. Given, Magnetic field (B) = 3000 G = 3000 × 10– 4 T = 0.3 T
• Length of the rectangular loop (l) = 10 cm
• Width of the rectangular loop (b) = 5 cm
• Area of the rectangular loop
• A = l × b
• = 10 × 5 = 50 cm2
• = 50 × 10– 4 m2
• Current in the loop (I) = 12 A
• Consider anti-clockwise direction of the current as positive and vice-versa :

$$\text{(i)}\space\text{Torque}\space (\vec{\tau})=\text{I}(\vec{A})×(\vec{B})\\\text{From figure, area vector}(\vec{A})\text{is normal to the y-z plane and B is directed along the z-axis.}\\\therefore\space\tau=12 × (50 ×10^{-4})\hat{i}×0.3×\hat{k}\\=-1.8×10^{-2}\hat{j}\space\text{Nm}\\\text{Net torque is 1.8×10}^{-2}\hat{j}\text{\space Nm\space along the negative y-direction.}\\\text{The force on the loop is zero because the angle between A and B is zero.}\\\text{(ii) Same as (i)}\\\text{(iii) Torque}\space(\vec{\tau})=\text{I}(\vec{A})×(\vec{B})\\\text{From the given figure, it can be observed that A is normal to the x-z plane and B is directed along the z-axis.}\\\therefore\space\tau=-12×(50×10^{-4})\hat{j}×0.3\hat{k}\\\lbrack\because\space\hat{j}×\hat{k}=\hat{i}\rbrack\\=-1.8×10^{-2}\hat{i}\space\text{Nm}\\\text{Torque is}\space1.8×10^{-2}\text{Nm along the negative x direction and the force is zero.}\\\text{(iv) Magnitude of torque is given as :} |\tau|=\text{IAB}\\=12×50×10^{-4}×0.3\\=1.8×10^{-2}\space\text{Nm}\\\text{The torque is}\space1.8×10^{-2}\text{Nm at an angle of 240° with positive x direction. The force is zero.}\\\text{(V)}\space\text{Torque},\tau=\text{I}\vec{A}×\vec{B}\\=(50×10^{-4}×12)\hat{k}×0.3×\hat{k}\\\lbrack\because\space\hat{k}×\hat{k}=0\rbrack\\=0\\\text{Therefore, the torque is zero. The force is also zero.}\\\text{(vi) Torque,}\space\tau=\text{I}\vec{A}×\vec{B}\\=(50×10^{-4}×12)\hat{k}×0.3\hat{k}\\\lbrack\because \hat{k}×\hat{k}=0\rbrack\\=0\\\text{The torque is zero. The force is also zero.}$$

$$\text{In case (v),}\space\text{I}\vec{A}\space\text{and}\space\vec{B}\text{is in same direction and the angle between them is zero.}\\\text{ If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.}\\\text{In case (vi),I}\vec{A}\space\text{and}\space\vec{B}\text{is in opposite direction.}\\\text{The angle between them is 180°. If displaced it does not come back to its original position. Hence, its equilibrium is unstable.}$$

Q. A galvanometer coil has a resistance of 12 W and the metre shows full scale deflection for a current of 3 mA. How will you convert the
metre into a voltmeter of range 0 to 18 V ?

Ans. Given, Resistance of galvanometer coil, G = 12 Ω Current for full scale deflection,

Current for full scale deflection,

Ig = 3 mA = 3 × 10– 3 A

Range of the voltmeter needs to be converted from 0 to 18 V.

Range of the voltmeter needs to be converted from 0 to 18 V.

$$\therefore\text{V}=18\text{V}\\\text{Consider a resistor of resistance R is connected in series with the galvanometer to convert it into a voltmeter. This resistance is given by}\\\text{R}=\frac{V}{I_s}-G\\=\frac{18}{3 ×10^{-3}}-12\\=5988\spaceΩ\\\text{Therefore, a resistor of resistance 5988 Ω is required to be connected in series with the galvanometer.}$$

Q. If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig.) which way would the
Lorentz force be for (a) an electron (negative charge). (b) a proton (positive charge). Ans. The velocity v of particle is in the direction of x-axis, while B, the magnetic field is in the
direction of y-axis, so v × B is along the z-axis (screw rule or right-hand thumb rule).
(i) Thus, for electron it will be along – z-axis.
(ii) For a positive charge (proton) the force is along + z-axis.