Question

At some location on earth the horizontal component of earth's magnetic field is $18\times {10}^{-6}\text{T}$. At this location, magnetic needle of length $0.12\text{m}$ and pole strength $1.8\text{Am}$ is suspended from its mid-point using a thread, it makes ${45}^{\circ }$ angle with horizontal in equilibrium.To keep this needle horizontal,the vertical force that should be applied at one of its ends is

• A. $3.6\times {10}^{-5}\text{N}$
• B. $1.8\times {10}^{-5}\text{N}$
• C. $1.3\times {10}^{-5}\text{N}$
• D. $6.5\times {10}^{-5}\text{N}$

Answer 1

The correct option is D $6.5\times {10}^{-5}\text{N}$

At ${45}^{\circ },B_v=B_H$


Using Torque equation,
$\tau =M{B}_v\sin \theta =F\frac{l}{2}\sin \theta$


$M{B}_v\sin {45}^{\circ }=F\frac{l}{2}\sin {45}^{\circ }$
$F=\frac{2M{B}_v}{l}=2\times 1.8\times 18\times {10}^{-6}=6.5\times {10}^{-5}\text{N}$.