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← JEE Main 2026 · 6 Apr · Shift 2
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JEE Main 2026 · 6 Apr · Shift 2, Physics question 39

A square loop of side 2 cm is placed in a time varying magnetic field with magnitude as B=0.4sin⁡(300t)B=0.4\sin(300t) Tesla. The normal to the plane of loop makes an angle of 60° with the field. The maximum induced emf produced in the loop is ________ mV.

Choose one option
Answer and worked solution

Official answer: D

Flux. Φ=BAcos⁡60∘\Phi=BA\cos60^\circ.

Maximum emf.
εmax⁡=B0ωAcos⁡60∘=0.4×300×4×10−4×0.5=24 mV.\varepsilon_{\max}=B_0\omega A\cos60^\circ=0.4\times300\times4\times10^{-4}\times0.5=24\text{ mV}.

(D)