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NCERT Class XII Chapter
Moving Charges and Magnetism
Questions With Solutions

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Question : 24
Total: 28
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 figure? What is the force on each case? Which case corresponds to stable equilibrium?

Solution:  
(a) Let us detail each case separately,

Dipole moment is along +x direction
→
M
 = 10 × 5 × 10−4×12
^
i
 = 0.06
^
i
→
B
 = 3000 × 10−4
^
k
Torque
→
τ
 =
→
M
×
→
B
 = - 1.8 × 10−2
^
j
 N m
So, torque is 1.8 × 10–2 N- m along –y direction net force on the coil is zero coil not in equilibrium.

(b) Dipole moment is along +x direction
→
M
 = 0.06
^
i
 ,
→
B
 = 0.3
^
k
Torque
→
τ
 =
→
M
×
→
B
 = - 1.8 × 10−2
^
j
 N m
So, torque is 1.8 × 10–2 N- m along –y direction
Net force on the coil is zero, coil is not in equilibrium
(c) Dipole moment is along –y direction

→
M
 = - 0.06
^
j
 , B = 0.3
^
k
Torque
→
τ
 =
→
M
×
→
B
 = - 1.8 × 10−2
^
j
 N m
So, torque is 1.8 × 10–2 N- m along - x direction.
Net force on the coil is zero, coil is not in equilibrium.
(d) Dipole moment is at an angle 150° with the +x direction.

Torque τ = MB sinθ
τ = 1.8 × 10−2
^
j
 sin
π
2
 = 1.8 × 10−2 N m
At an angle 240° with the +x direction net force on the coil is zero, coil is not in equilibrium.
(e) Dipole moment is along +z direction.

→
τ
 =
→
M
×
→
B
 = - 1.8 × 10−2
^
k
×
^
k
 = 0
Potential energy U = –
→
M
.
→
B
U = - 1.8 × 10−2 (
^
k
.
^
k
)
= - 1.8 × 10−2 J
Negative energy shows equilibrium is stable.
(f) Dipole moment is along – z direction

→
τ
 =
→
M
×
→
B
 = - 1.8 × 10−2−
^
k
×
^
k
 = 0
P.E, U = –MB
= + 1.8 × 10−2 J
Positive energy shows equilibrium is unstable.
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