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JEE PHYSICS MCQ: MAGNETIC EFFECT OF CURRENT

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Application +4 | -1
A +6μC+6 \mu \mathrm{C} point charge is moving at a constant velocity of 8×106 ms−18 \times 10^6 \mathrm{~ms}^{-1} in the +y+y direction, relative to a reference frame. At the instant when the point charge is at the origin of this reference frame, what is the magnetic field vector it produces at the following points?
(a) x=0.500 m,y=0,z=0x=0.500 \mathrm{~m}, y=0, z=0, and
(b) x=0,y=−0.500 m,z=0x=0, y=-0.500 \mathrm{~m}, z=0.
View answer and explanation
Correct option: A
−(1.92×10−5 T)k^;0-\left(1.92 \times 10^{-5} \mathrm{~T}\right) \hat{k}; 0
Option A (−(1.92×10−5 T)k^;0-(1.92 \times 10^{-5} \mathrm{~T}) \hat{{k}}; 0) is correct because it correctly calculates the magnetic field using the Biot-Savart law for a moving point charge.

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