JEE Mains 2026 APR 5th Shift 2 Q38 KE Change in Crossed E and B Fields Option 3 #jeemain2026

追加: 2026-04-29

[00:00:00]G means 5th April 2026. Shift 2.

[00:00:04]Question 38.

[00:00:06]A charged particle of charge Q and mass M is projected from the origin with velocity V / <unk>2 in the X direction plus V over <unk>2 in the Y direction.

[00:00:15]There is a uniform magnetic field V in the Z direction and a space varying electric field E to the minus lambda X in the X direction within 0 less than or equal to X less than or equal to L. Let us visualize the setup.

[00:00:30]A particle of charge Q and mass M is projected from origin with initial velocity V /<unk> to Xhat plus V /<unk> to Yhat. A uniform magnetic field V zhat hat and an electric field E not E to the minus lambda Xhat exist in the region 0 to L. After the X coordinate changes from 0 to L, find the change in kinetic energy. The key concept is the work energy theorem. The magnetic force is always perpendicular to the velocity. So it does zero work. The change in kinetic energy equals the work done by the electric field only.

[00:01:05]Step one, the work done by the electric field is the integral from 0 to L of Q * E the minus lambda X dx. Since the electric field is in the x direction and we integrate over x, this is a straightforward one-dimensional integral.

[00:01:20]Step two, evaluating the integral. The integral of e to the minus lambda x is - 1 / lambda * e the minus lambda x.

[00:01:28]Evaluating from 0 to l gives 1 / lambda * 1 - e to the minus lambda l.

[00:01:33]Multiplying by q not the change in kinetic energy is q / lambda * 1 - e to the minus lambda l. The correct answer is option 1. Q / lambda * 1 - e to the minus lambda.

[00:01:47]G tip. The magnetic force never does work because it is always perpendicular to the velocity. This is a fundamental property. Whenever a problem involves both electric and magnetic fields, only the electric field contributes to change in kinetic energy. This simplifies the integral significantly.

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