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JEE Main 2020 (4 Sep, Shift 1) · Field of a Magnet & Magnet in a Uniform Field

Field of a Magnet & Magnet in a Uniform Field question from JEE Main 2020 (4 Sep, Shift 1)

Q15JEE Main 2020 · 4 SEP · SHIFT 1Medium

Field of a Magnet & Magnet in a Uniform Field

A small bar magnet placed with its axis

at 30° with an external field of 0.06 T

experiences a torque of 0.018 Nm. The

minimum work required to rotate it from its

stable to unstable equilibrium position is:

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Answer: (c)

  1. τ=MBsin⁡30∘⇒MB=0.0180.5=0.036 J\tau = MB\sin 30^\circ \Rightarrow MB = \frac{0.018}{0.5} = 0.036\ \text{J}
  2. Stable (0∘0^\circ) to unstable (180∘180^\circ): W=MB(cos⁡0∘−cos⁡180∘)=2MBW = MB(\cos 0^\circ - \cos 180^\circ) = 2MB
  3. W=2×0.036=0.072 JW = 2 \times 0.036 = 0.072\ \text{J}
  4. So the answer is (c) 7.2×10−2 J7.2 \times 10^{-2}\ \text{J}.
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