Question

The torgue required to keep a magnet of length $20cm$ at ${30}^o$ to a uniform field is $2{\times }^{-5}N-m$. The magnetic force on each pole is

• A. $2{\times }^{-4}N$
• B. $4\times {10}^{-4}N$
• C. $1\times {10}^{-3}N$
• D. $2\times {10}^{-6}N$

Answer 1

The correct option is A $2{\times }^{-4}N$

We have,

Length of magnetic field is $0.20m$, Magnetic field $2\times {10}^{-5}Am$

So, $\tau =mB\sin {30}^0=0.20\times 2\times {10}^{-5}\times \frac{1}{2}=2\times {10}^{-4}N$