JEE Main 2022 · 27 Jun · Shift 2 · Physics Q59

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PhysicsInteger+4 / -1Question 59 of 90

A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is $4\times10^{-3}$ T and the angle of dip is 45°. The emf induced in the rod is ______ mV.

Your answer

Answer and worked solution

Official answer: 16

Vertical field (dip 45°): $B_V=B_H\tan45°=4\times10^{-3}$ T, which the eastward-moving N-S rod cuts.

emf. $B_Vvl=4\times10^{-3}\times20\times0.2=16$ mV.

Answer: 16

Topics:Motional EMF Earth's magnetism

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JEE Main 2022 · 27 Jun · Shift 2, Physics question 59