AP EAMCET · PHYSICS · Magnetic Effects of Current
The work done in rotating a bar magnet which is initially in the direction of a uniform magnetic field through \(45^{\circ}\) is W. The additional work to be done to rotate the magnet further through \(15^{\circ}\) is
- A \(\frac{\mathrm{W}}{\sqrt{2}}\)
- B \(\frac{\mathrm{W}}{2}\)
- C \(\mathrm{W} \sqrt{2}\)
- D 2 W
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{W}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(W = MB(\cos 0^{\circ} - \cos 45^{\circ})\) \(W = MB(1 - \frac{1}{\sqrt{2}})\) \(W' = MB(\cos 45^{\circ} - \cos 60^{\circ})\) \(W' = MB(\frac{1}{\sqrt{2}} - \frac{1}{2})\) \(W' = \frac{W}{1 - \frac{1}{\sqrt{2}}} (\frac{1}{\sqrt{2}} - \frac{1}{2})\)…
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