Question

If a magnet of length $10cm$ and pole strength $40A-m$ is placed at an angle of ${45}^{}$ in a uniform induction field of intensity $2\times {10}^{-4}T$. The couple acting on it is

• A. $0.5656\times {10}^{-3}N-m$
• B. $0.565\times {10}^{-3}N-m$
• C. $0.756\times {10}^{-4}N-m$
• D. None of the above

Answer 1

Answer: (B)

$0.565\times {10}^{-3}N-m$

Given $l=10cm$

Pole strength = $40A-m$

M $=40\times 0.1=4A-m^2$ $B=2\times {10}^{-4}T$

$\tau =M\times Bsin\theta =4\times 2\times sin{45}^o\times {10}^{-4}=8\times \frac{1}{\sqrt{2}}\times {10}^{-4}=5.656\times {10}^{-4}N-m$