Question

A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $60^\circ $ and one of the fields has a magnitude of $1.2\times 10^{- 2}T$ . If the dipole comes to stable equilibrium at an angle of $15^\circ $ with this field, what is the magnitude of the other field?

• A. $4.4\times 10^{- 3}$
• B. $4.4\times 10^{- 4}$
• C. $4.8\times 10^{- 3}$
• D. $4.4\times 10^{3}$

Answer 1

Answer: (A)

Let one of the magnetic fields is B₁ and other is B₂. Angle between B₁ and B₂ is 60^o.
Given, B₁ = 1.2 x 10^-2 T
Dipole is in equilibrium at an angle 15^o from B₁ or 60^o - 15^o = 45^o from B₂
Torque on dipole due to magnetic field B₁
$\tau_{1} = \text{M} \times \text{B}_{1} \text{ sin} 1 5^{\text{o}}$ ......(i)

where, M is the magnetic moment. Torque on dipole due to magnetic field B₂
$\tau_{2} = \text{M} \times \text{B}_{2} \text{ sin} 4 5^{\text{o}}$ .....(ii)
As dipole is in the equilibrium, then
$\tau_{1} = \tau_{2}$
M X B₁ sin 15^o = M x B₂ sin 45^o [From Eqs. (i) and (ii)]
$\frac{1 \text{.} 2 \times 1 0^{- 2} \times \text{sin } 1 5^{\text{o}}}{\text{sin } 4 5^{\text{o}}} = \text{B}_{2}$
$\text{B}_{2} = \frac{1 \text{.} 2 \times 1 0^{- 2} \times 0 \text{.} 2 5 8 8}{0 \text{.} 7 0 7 1} = 4 \text{.} 4 \times 1 0^{- 3} \text{ T}$
Thus, the magnitude of the other field is 4.4. x 10^-3 T.