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MHT CET 2025 (26 Apr, Shift 1) · Field of a Magnet & Magnet in a Uniform Field

Field of a Magnet & Magnet in a Uniform Field question from MHT CET 2025 (26 Apr, Shift 1)

Q4MHT CET 2025 · 26 APR · SHIFT 1Medium

Field of a Magnet & Magnet in a Uniform Field

The work done in turning a magnet of magnetic moment ' MM ' by an angle of 90∘90^\circ from the meridian is ' nn ' times the corresponding Work done to turn it through an angle of 60∘60^\circ where the value of ' nn ' is (cos⁡90∘=0,cos⁡60∘=0.5)(\cos 90^\circ = 0, \cos 60^\circ = 0.5)

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Answer: (b)

  1. W=MB(1−cos⁡θ)W = MB(1 - \cos\theta) starting from the meridian.
  2. W90=MB(1−0)=MBW_{90} = MB(1 - 0) = MB, W60=MB(1−0.5)=MB2W_{60} = MB(1 - 0.5) = \dfrac{MB}{2}
  3. n=W90W60=2n = \dfrac{W_{90}}{W_{60}} = 2
  4. So the answer is (b) 2.
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