First slide

Magnetism

Question

A magnetic dipole is under the influence of two magnetic fields the angle between the two field directions is $60^{0}$ and one of the fields has a magnitude of $1.2 X 10^{- 2} T$ . If the dipole comes to stable equilibrium at an angle of $15 °$ with this field, what is the magnitude of the other field?

Moderate

A

$2.39 \times 10^{- 3} T$
B

$3.39 \times 10^{- 3} T$
C

$4.39 \times 10^{- 3} T$
D

$5.39 \times 10^{- 3} T$

Solution

Given $B_{1} = 1.2 \times 10^{- 2} T$
Inclination of dipole with $B_{1} is θ_{1} = 15 °$
Therefore Inclination of dipole with $B_{2} is θ_{2} = 60 ° - 15 ° = 45 °$

As the dipole is in equilibrium the torque on the dipole due to the two fields are equal and opposite. If M is the magnetic dipole moment of the dipole, then
$\begin{array}{l} ∴ M B_{1} \sin θ_{1} = M B_{2} \sin θ_{2} \\ \Rightarrow B_{2} = \frac{B_{1} \sin θ_{1}}{\sin θ_{2}} \\ \Rightarrow B_{2} = \frac{1.2 \times 10^{- 2} \times \sin 15 °}{\sin 45^{o}} = \frac{1.2 \times 10^{- 2} \times 0.2588}{0.707} \\ \Rightarrow B_{2} = 4.39 \times 10^{- 3} T \end{array}$

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