A coil having 100 turns, area of 5 × 10^ -3 ~m^2 , carrying current of 1 ~mA is placed in uniform magnetic field of 0.20 ~T such a way that plane of coil is perpendicular to the magnetic field. The work done in turning

Correct answer: 0 aligned & W=Δ U=U_f-U_i & W=(- μ . B )_f-(- μ . B )_i & =0+( μ · B )_i & =(100 × 5 × 10^ -3 × 1 × 10^ -3 ) × 0.2 ~J & =1 × 10^ -4 ~J=100 μ J aligned

JEE Main 20246 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Evening Shift
Subject: Physics
Chapter: Magnetic Effects of Current

JEE Main 2024 · 6 April · Evening Shift · Physics · Magnetic Effects of Current

A coil having 100 turns, area of 5 × 10^ -3 ~m^2 , carrying current...

A coil having 100 turns, area of

5 \times 10^{-3} \mathrm{~m}^2

, carrying current of

1 \mathrm{~mA}

is placed in uniform magnetic field of

0.20 \mathrm{~T}

such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through

90^{\circ}

is _______

\mu \mathrm{J}

Correct answer

Step-by-step solution

\begin{aligned} & W=\Delta U=U_f-U_i \\ & W=(-\vec{\mu} . \vec{B})_f-(-\vec{\mu} . \vec{B})_i \\ & =0+(\vec{\mu} \cdot \vec{B})_i \\ & =\left(100 \times 5 \times 10^{-3} \times 1 \times 10^{-3}\right) \times 0.2 \mathrm{~J} \\ & =1 \times 10^{-4} \mathrm{~J}=100 \mu \mathrm{J}\end{aligned}

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