AP EAMCET · PHYSICS · Magnetic Effects of Current
A magnetic needle lying parallel to a magnetic field requires \(W\) units of work to turn it through \(60^{\circ}\). The torque required to maintain the needle in this position will be
- A \(\sqrt{3} W\)
- B \(W\)
- C \(\frac{\sqrt{3}}{2} W\)
- D \(2 \mathrm{~W}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{3} W\)
Step-by-step Solution
Detailed explanation
According to the question, work done required to rotate magnetic needle from \(\theta_1=0^{\circ}\) to \(\theta_2=60^{\circ}\). \(W=M B\left(\cos \theta_1-\cos \theta_2\right)\)…
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