A short bar magnet placed with its axis at \(30°\) with an external field of \(800\) G experiences a torque of \(0.016\) Nm. What is the work done in moving it from its most stable to most unstable position?
Answer
\(0\mathrm{\ldotp }064J\)
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Solution
Torque, \(\tau =mBsin\theta\)
\(m=\dfrac{\mathit{\tau }}{Bsin\mathit{\theta }}=\dfrac{0\ldotp 016}{800\times {10}^{-4}\times \sin {30}^{o}}\) [ As 1G=10-4T ]
\(m=0\ldotp 4A{m}^{2}\)
Work done to rotate it from stable to unstable is given by:
\(\therefore \;W=\int \nolimits_{0{}^{\circ }}^{180{}^{\circ }}\;[-m\;B\sin \theta ]\)
\(=−{m}{B}\left. \left[\cos 180{}^{\circ }−\cos 0{}^{\circ }\right]\right.\)
\(=−mB\left. \left[-1−1\right]\right.\)
\(=2mB\)
\(=2\times 0\ldotp 40\times 800\times {10}^{−4}=0\ldotp 064J\)
\(m=\dfrac{\mathit{\tau }}{Bsin\mathit{\theta }}=\dfrac{0\ldotp 016}{800\times {10}^{-4}\times \sin {30}^{o}}\) [ As 1G=10-4T ]
\(m=0\ldotp 4A{m}^{2}\)
Work done to rotate it from stable to unstable is given by:
\(\therefore \;W=\int \nolimits_{0{}^{\circ }}^{180{}^{\circ }}\;[-m\;B\sin \theta ]\)
\(=−{m}{B}\left. \left[\cos 180{}^{\circ }−\cos 0{}^{\circ }\right]\right.\)
\(=−mB\left. \left[-1−1\right]\right.\)
\(=2mB\)
\(=2\times 0\ldotp 40\times 800\times {10}^{−4}=0\ldotp 064J\)
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