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A short bar magnet placed with its axis at \(30\degree\) with an external field of \(800\; G\)  experiences a torque of \(0.016\;Nm.\)  The bar magnet is replaced by a solenoid of cross-sectional  are \(2 \times 10^{-4}\;m^2\)  and \(1000\) turns, but of the same magnetic moment. Determine the current flows through the solenoid.

Answer

2A
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Solution

We know \(\tau=m B \sin \theta . \theta=30^{\circ},\) 
hence \(\sin \theta=\frac 1 2\)
Thus, \(0.016=m \times\left(800 \times 10^{-4} T \right) \times(\frac 1 2)\)
\(m=\frac{160 \times 2 }{ 800}=0.40 Am ^{2}\)
The bar magnet is replaced by a solenoid of cross-sectional area  \(2\times 10^{-4}\;m^2\)  and \(1000\) turns, but of the same magnetic moment. Determine the current flows through the solenoid.
We need to apply the formula for the magnetic moment:
 \({\therefore}\;m=NI A\) 
 \(0.40=1000\times I\times 2\times 10^{-4}\) 
 \(\Rightarrow I=2\;A\) 
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