Question

Question 4.24: 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. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-171

Answer 1

(a) magnetic field strength ,B =3000G = 0.3T
length of rectangular loop , l = 10cm = 0.1m
wodth of rectangular loop, b = 5 cm = 0.05 m
so, area of loop, A = l × b = 0.005 m²
current in the loop, I = 12A
now $\vec{\tau}=I(\vec{A}\times\vec{B})$
here, A = 0.005 i
B = 0.3 k
so, $\tau=12(0.005 i \times 0.3k)$
= - 1.8 × 10^-2 Nm j
the force on the loop is zero. because external magnetic field is uniform.

(b) This case is similar to A, hence the results will also be same. There is no change in any vectors.

(c) T = IA × B
We observe that A is perpendicular to x-z plane and B is along z axis.
⇒ T = -12 × (5 × 10^-3) j× 0.3k
⇒ T = -1.8 × 10^-2 Nm i
and the force is zero.

(d) Magnitude of torque,
|T| = IA × B
⇒ |T| = 12 × (50 × 10-4)m2 × 0.3T
⇒ |T| = 1.8 × 10-2 Nm
This torque is makes 240° with the positive x direction. The force again is Zero.[ because sin240° = 0]

(e) T = IA × B
 = 12 × (50 × 10-4) m2 × 0.3 T
 = 0
Since cross section and magnetic field are in same direction net torque is zero. Net force is also Zero.

(f) torque =12 (5 × 10^-3) k × 0.3k
= 0
and force is also zero .

Stable equilibrium:
In case E, the angle between A and B is zero. If we displace the wire, it will come back in this position, hence it is the stable equilibrium condition.
Unstable equilibrium:
In case F, the angle between A and B is 180°. If the wire is displaced, it will not come back in this position so we can conclude that it is the case for unstable equilibrium