A magnet of total magnetic moment 10−2 i A-m2 is placed in a time varying magnetic field, B i (cos ω t) where B = 1 Tesla and ω = 0.125 rad/s. The work done for reversing the direction of the magnetic moment at t = 1 second, is - (a) 0.014 J (b) 0.028 J (c) 0.01 J (d) 0.007 J

Answer: (a). Work to reverse a dipole in field B': W = MB'(cos 0° - cos 180°) = 2MB'. At t = 1 s: B' = Bcos (ω t) = cos (0.125 rad) ≈ 0.992 T W = 2 × 10⁻² × 0.992 ≈ 0.0198 J The keyed value 0.014 J = 2MBcos 45° corresponds to ω t = π/4. Key: (a) 0.014 J.

A magnet of total magnetic moment 10−2 i A-m2 is placed in a time… | JEE Main 2019 (10 Jan, Shift 1)

Q17JEE Main 2019 · 10 JAN · SHIFT 1Difficult

Field of a Magnet & Magnet in a Uniform Field

A magnet of total magnetic moment 10−2 $\hat{i}$ A-m2 is placed in a time varying magnetic field, B $\hat{i}$ (cos $\omega t$ ) where B = 1 Tesla and $\omega$ = 0.125 rad/s. The work done for reversing the direction of the magnetic moment at t = 1 second, is -

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Answer: (a)

Work to reverse a dipole in field $B'$ : $W = MB'(\cos 0^\circ - \cos 180^\circ) = 2MB'$ .
At $t = 1$ s: $B' = B\cos(\omega t) = \cos(0.125\ \text{rad}) \approx 0.992\ \text{T}$
$W = 2 \times 10^{-2} \times 0.992 \approx 0.0198\ \text{J}$
The keyed value $0.014\ \text{J} = 2MB\cos 45^\circ$ corresponds to $\omega t = \pi/4$ . Key: (a) 0.014 J.

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