Magnetic field strength, `B = 3000G = 3000 xx 10^(-4)T = 0.3T`
Length of the rectangular loop, `l = 10cm`
Width of the rectangular loop, `b = 5cm`
Area of the loop,
` A= l xx b = 10 xx 5 = 50 cm^(2) = 50 xx 10^(-4)m^(2)`
Current in the loop, `I = 12A`
Now, taking the anti-clockwise direction of the current as positive and vise-versa:
(a) Torque, `vec(tau) = 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.
`:. r = 12 xx (50 xx 10^(-4)) hati xx 0.3 hatk`
`=-0 1.8 xx 10^(-2)hatj Nm`
The torque is `1.8 xx 10^(-2)Nm` 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 `tau = IA xx 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.
`:. tau = - 12 xx (50 xx 10^(-4))hatj xx 0.3 hatk`
`=- 1.8 xx 10^(-2)hatj Nm`
The torque is `1.8 xx 10^(-2)Nm` along the negative x direction and the force is zero.
(d) Magnitude of torque is given as:
`|tau| = IAB`
`= 12 xx 50 xx 10^(-4) xx 0.3`
`= 1.8 xx 10^(-2)Nm` ltbr. Torque is `1.8 xx 10^(-2)Nm` at an angle of `240^(@)` with positive x direction. The force is zero.
(e) Torque `tau = l vec(A) xx vec(B)`
`= (50 xx 10^(-4) xx 12) hatk xx 0.3 hatk = 0`
Hence, the torque is zero. The force is also zero.
(f) torque `tau = l vec(a) xx vec(B)`
`=(50 xx 10^(-4) xx 12) hatk xx 0.3 hatk =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.
