A uniform magnetic field of `3000G` is established along the positive z-direction. A rectangular loop of sides `10cm` and `5cm` carries a current `12A`. What is the torque on the loop in the different cases shown in the figure. What is the force on each case? Which case corresponds to stable equilibrium?

Explanation

(a) Torque on a loop of area `vecA`, carrying a current I placed in an external magnetic field `vecB` is given by:
`vec tau = I (vecA xx vecB)`
Here, ` B = 3,000 G = 3,000 xx 10^(-4) T = 0.3 T`
Since B is along the positive Z-axis,
` vecB = 0.3 hatk T`
` A = 10 xx 5 = 50 cm^(2) = 50/(100 xx 100) m^(2) = 5 xx 10^(-3) m^(2)`
Since current flowing in the loop is in the anticlockwise direction and the loop is lying in the Z-Y plane, the area veclor of the loup will be along positive X-axis.
`vecA = 5 xx 10^(-3) hati`
` I = 12 A`
`rArr vec tau = I (vecA xx vecB) = 12 [(5 xx 10^(-3) hati) xx (0.3 hatk)]`
` = - 1.8 xx 10^(-2) hatj N m`
Torque acting on the rod will be along negative Y-axis.
(b) Since `vecB` is along positive Z-axis,
` vecB = 0.3 hatk T`
Since current flowing in the loop is in the anticlockwise direction and the loop is lying in the Z-Y plane, the area vector of the loop will be along positive X-axis.
` vecA = 5 xx 10^(-3) hati m^(2)`
`rArr" " vec tau = I (vec A xx vec B) = 12 [(5 xx 10^(-3) hati) xx (0.3 hat k)]`
` = - 1.8 xx 10^(-2) hat j` N m
Torque acting on the rod will be along negative Y-axis.
(c) Since `vecB` is along positive Z - axis ,
` vecB = 0.3 hat k`
Since current flowing in the loop is in the clockwise direction and the loop is lying in the X-Z plane, the area vector of the loop will be along negative Y axis
` vec A = 5 xx 10^(-3) (- hat j ) m^(3)`
` rArr " " hat tau = I ( vecA xx vec B) = 12 [(-5 xx 10^(-3) hat j) xx (0.3 hat k)]`
` = - 1.8 xx 10^(-2) hat i ` N m
Torque acting on the rod will be along negative X - axis.
(d) The current is flowing in the anticlockwise direction and the loop is oriented at an angle of `30^(@)` from the Z-Y plane. Thus, area vectors is given by:
`vecA = A sin 30^(@) (- hatj ) + A cos 30^(@) ( hati)`
` = 5 xx 10^(-3) [sqrt3/2 hati - 1/2 hatj] m^(2)`
` vecB = 0.3 hatk T`
`rArr hat tau = I (vecA xx vecB) = 12 [5 xx 10^(-3) (sqrt3/2 hati - 1/2 hatj) xx (0.3 hat k)]`
` = -0.9 xx 10^(-2) (hati + sqrt3 hatj ) N m`
` |vec tau| = 1.8 xx 10^(-2) N m`
Torque acting on the rod is at an angle of `240^(@)` with the positive X direction.
(e) `vecB = 0.3 hat k T`
Since current flowing in the loop is in the anticlockwise direction and the loop is lying in the X-Y plane, the area vector of the loop will be along positive Z - axis
`vecA = 5 xx 10^(-3) hat k`
` rArr" " vec tau = I (vec A xx vec B) = 12 [(5 xx 10^(-3) hat k) xx (0.3 hat k)] = 0 `
(f) ` vec B = 0.3 hat k`
Since current flowing in the loop is in the clockwise direction and the loop is lying in the X-Y plane, the area vector of the loop will be along negative Z-axis.
` vecA = - 5 xx 10^(-3) hat k`
`rArr vec tau = I (vec A xx vec B) = 12 [(-5 xx 10^(-3) hat k) xx (0.3 hat k)] = 0 `
As the loop is closed and placed in a uniform magnetic field, force is zero in each case. Case (e) corresponds to a stable equilibrium because in this case, force and torque acting on the loop are both zero. Also, the magnetic moment of the loop given by `M = I vec A` is along the direction of `vec B`. Case (f) corresponds to unstable equilibrium because the magnetic moment in this case is opposite to the direction of B.