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Torque on current carrying loop - class-XII
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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
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A
$2\times ^{-4}N$
💡 Explanation:
We have,
Length of magnetic field is $0.20m$, Magnetic field $2\times10^{-5}Am$
So, $\tau=mB\sin30^0=0.20\times2\times10^{-5}\times\dfrac{1}{2}=2\times10^{-4}N$