JEE Main 2019 — Electromagnetic Induction Question with Solution
From: JEE Main 2019 (Online) 10th January Morning Slot
Question
Choose an option
Show full solutionCorrect option: B
Step-by-step explanation
We can apply Faraday's law of electromagnetic induction to solve this problem. Faraday's law states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.
The cube is moving through a magnetic field, so it's behaving like a conductor moving through a magnetic field. The induced emf or voltage can be calculated by using the formula:
emf = B v d
Where:
B is the magnetic field strength,
v is the velocity of the conductor, and
d is the length of the conductor perpendicular to the direction of motion and magnetic field.
In this case, the cube is moving in the y-direction and the magnetic field is in the z-direction. So, the faces perpendicular to the x-axis are involved. The length of the conductor (d) perpendicular to the motion and magnetic field is the edge length of the cube, which is 2 cm or 0.02 m.
So, plugging the given values into the formula:
emf = 0.1 T 6 m/s 0.02 m = 0.012 V = 12 mV
So, the potential difference between the two faces of the cube perpendicular to the x-axis is 12 mV.
Therefore, Option B is correct.
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This is a previous-year question from JEE Main 2019, covering the Electromagnetic Induction chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.