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TS EAMCET · Physics · Magnetic Effects of Current

A short bar magnet of magnetic moment \(2.5 \mathrm{Am}^2\) is kept in a uniform magnetic field of \(4 \times 10^{-5} \mathrm{~T}\). The work done in moving the magnet from its most stable position to most unstable position is

  1. A \(40 \times 10^{-5} \mathrm{~J}\)
  2. B \(25 \times 10^{-5} \mathrm{~J}\)
  3. C \(10 \times 10^{-5} \mathrm{~J}\)
  4. D \(20 \times 10^{-5} \mathrm{~J}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(20 \times 10^{-5} \mathrm{~J}\)

Step-by-step Solution

Detailed explanation

\(W = -MB(\cos\theta_2 - \cos\theta_1)\) \(W = -(2.5)(4 \times 10^{-5})(\cos 180^\circ - \cos 0^\circ)\) \(W = -(10 \times 10^{-5})(-1 - 1)\) \(W = -(10 \times 10^{-5})(-2)\) \(W = 20 \times 10^{-5} \, \mathrm{J}\)