TS EAMCET · Physics · Magnetic Effects of Current
A circular coil of 10 turns and radius \(10 \mathrm{~cm}\) is placed in a uniform magnetic field of \(0.1 \mathrm{~T}\) normal to the plane of the coil. If the current in the coil is \(5 \mathrm{~A}\), then the magnitude of the torque on the coil is
- A \(500 \pi \mathrm{N}-\mathrm{m}\)
- B \(0.05 \pi \mathrm{N}-\mathrm{m}\)
- C \(0.005 \pi \mathrm{N}-\mathrm{m}\)
- D Zero
Answer & Solution
Correct Answer
(D) Zero
Step-by-step Solution
Detailed explanation
Magnitude of torque on the coil is given as \(\tau=|\mathbf{m} \times \mathbf{B}| \Rightarrow \tau=N I A B \sin \theta\) where, \(\theta\) is the angle between axis of loop (normal to the loop) and direction of magnetic field. Here, \(\theta=0 \quad\) So, \(\quad \tau=0\)
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