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Karnataka Board PUCPUC Science 2nd PUC Class 12

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 the loop in the

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Question

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 the loop in the different cases shown in fig.? What is the force on each case? Which case corresponds to stable equilibrium?

Numerical
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Solution

Magnetic field strength, 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 loop,

A = l × b = 10 × 5 = 50 cm2 = 50 × 10−4 m2

Current in the loop, I = 12 A

Now, taking the anti-clockwise direction of the current as positive and vice-versa:

(a) Torque, `vecτ = "I"vec"A" xx vec"B"`

From the given figure, it can be observed that A is normal to the y–z plane and B is directed along the z-axis.

∴ τ = `12 xx (50 xx 10^-4)hat"i" xx 0.3  hat"k"`

= `-1.8 xx 10^-2  hat"j"  "N m"`

The torque is 1.8 × 10−2 N m along the negative y-direction. The force on the loop is zero because the angle between A and B is zero.

(b) This case is similar to case (a). Hence, the answer is the same as (a).

(c) Torque, `τ = "I"vec"A" xx vec"B"`

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.

∴ τ = `-12 xx (50 xx 10^-4)hat"j" xx 0.3  hat"k"`

= `-1.8 xx 10^-2  hat"i"  "N m"`

The torque is 1.8 × 10−2 N m along the negative x-direction and the force is zero.

(d) Magnitude of torque is given as:

|τ| = IAB

= 12 × 50 × 10−4 × 0.3

= 1.8 × 10−2 N m

Torque is 1.8 × 10−2 N m at an angle of 240° with positive x-direction. The force is zero.

(e) Torque, `τ = "I"vec"A" xx vec"B"`

= `(50 xx 10^-4 xx 12)  hat"k" xx 0.3  hat"k"` 

= 0

Hence, the torque is zero. The force is also zero.

(f) Torque, `τ = "I"vec"A" xx vec"B"`

= `(50 xx 10^-4 xx 12)  hat"k" xx 0.3  hat"k"` 

= 0

Hence, the torque is zero. The force is also zero.

In case (e), the direction of `"I"vec"A"` and `vec"B"` is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.

Whereas, in case (f), the direction of `"I"vec"A"` and `vec"B"` is opposite. The angle between them is 180°. If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.

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Chapter 4: Moving Charges and Magnetism - Exercise [Page 171]

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NCERT Physics Part I and II [English] Class 12
Chapter 4 Moving Charges and Magnetism
Exercise | Q 4.24 | Page 171

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