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
${\tau }_1=\text{M}\times {\text{B}}_1\text{ sin}15^{\text{o}}$ ......(i)

where, M is the magnetic moment. Torque on dipole due to magnetic field B₂
${\tau }_2=\text{M}\times {\text{B}}_2\text{ sin}45^{\text{o}}$ .....(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)]
$\frac{1\text{.}2\times 10^{-2}\times \text{sin }15^{\text{o}}}{\text{sin }45^{\text{o}}}={\text{B}}_2$
${\text{B}}_2=\frac{1\text{.}2\times 10^{-2}\times 0\text{.}2588}{0\text{.}7071}=4\text{.}4\times 10^{-3}\text{ T}$
Thus, the magnitude of the other field is 4.4. x 10^-3 T.