Verified Solution • 2025–26 Session moderate

QJEEM24Pheb8e: JEE Main 2024 Physics - Class 12 JEEM24Pheb8e Solution: The magnetic moment of a bar magnet is . It is suspended in a…

The magnetic moment of a bar magnet is 0.5 Am20.5 \mathrm{~Am}^2 . It is suspended in a uniform magnetic field of 8×10−2 T8 \times 10^{-2} \mathrm{~T} . The work done in rotating it from its most stable to most unstable position is:

Multiple Choice Options

  • A 4×10−2 J4 \times 10^{-2} \mathrm{~J}
  • B 16×10−2 J16 \times 10^{-2} \mathrm{~J}
  • C 8×10−2 J8 \times 10^{-2} \mathrm{~J}
  • D Zero

Step-by-Step Verified Solution

Wamp;=∫0180dτ⋅dθamp;=mB(cos⁡0−cos⁡180)amp;=0.5×8×10−2(2)amp;=8×10−2 J\begin{aligned}W & =\int_0^{180} d \tau \cdot d \theta \\& =m B(\cos 0-\cos 180) \\& =0.5 \times 8 \times 10^{-2}(2) \\& =8 \times 10^{-2} \mathrm{~J}\end{aligned}

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