Question

A magnet is parallel to a uniform magnetic field. If it is rotated by $ 60{}^\circ $ , the work done is 0.8 J. How much work is done in moving it $ 30{}^\circ $ further?

• A. $ \text{0}\text{.8 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{7}}}\,\text{erg} $
• B. 4.0 J
• C. 8 J
• D. 0.8 erg

Answer 1

Answer: (A)

$ W=MB(\cos {{\theta }_{1}}-\cos {{\theta }_{2}}) $ When the magnet is rotated from $ {{0}^{o}} $ to $ {{60}^{o}} $ , then work done is $ 0.8\,\,J $ . $ 0.8=MB(\cos {{0}^{o}}-\cos {{60}^{o}})=\frac{MB}{2} $ $ \Rightarrow $ $ MB=1.6\,\,N-m $ In order to rotate the magnet through an angle of $ {{30}^{o}} $ , $ ie, $ from $ {{60}^{o}} $ to $ {{90}^{o}} $ , the work done is $ W=MB(\cos {{60}^{o}}-\cos {{90}^{o}})=MB\left( \frac{1}{2}-0 \right) $ $ =\frac{MB}{2}=\frac{1.6}{2}=0.8\,\,J=0.8\times {{10}^{7}}erg $