Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

243 of 276

Question 243 of 276

A coil having 100 turns, area of 5×10−3 m25 \times 10^{-3} \mathrm{~m}^2, carrying current of 1 mA1 \mathrm{~mA} is placed in uniform magnetic field of 0.20 T0.20 \mathrm{~T} such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through 90∘90^{\circ} is _______ μJ\mu \mathrm{J}.

Show answer and solution

Answer: 100

W=ΔU=Uf−UiW=(−μ⃗.B⃗)f−(−μ⃗.B⃗)i=0+(μ⃗⋅B⃗)i=(100×5×10−3×1×10−3)×0.2 J=1×10−4 J=100μJ\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}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Magnetic Effects & Magnetism (Magnetic Moment), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.