Magnetic Effects & Magnetism

JEE Main level, single correct, medium

148 of 276

Question 148 of 276

An electron is moving along +x+x direction with a velocity of 6×106ms−16\times {10}^{6}{ms}^{-1}. It enters a region of uniform electric field of 300V/cm300V/cm pointing along +y+y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be:

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Answer: (C) 5×10−3T,5\times {10}^{-3}T, along +z+z direction

q∨B=qEq∨B=qE
B=EV=300006×106=5×103TB=\dfrac{E}{V}=\dfrac{30000}{6\times {10}^{6}}=5\times {10}^{3}T
Magnetic force F=q(V⃗×B⃗)F=q(\vec{V}\times \vec{B})
So Ans.5×103T5\times {10}^{3}T ,along +Z direction

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