Magnetism and Matter

JEE Main 2024 (Online) 6th April Morning ShiftNumericalEasy📈 Topic in 7% papers+4

Ques. 1JEE Main 2024, 6 April (Morning Shift)

A circular coil having 200 turns, 2.5 10-4 m2 area and carrying 100 current is placed in a uniform magnetic field of 1 T. Initially the magnetic dipole moment ( M) was directed along B. Amount of work, required to rotate the coil through 90 from its initial orientation such that M becomes perpendicular to B, is J.

Answer

Solution

W = Uf - Ui = (-MB 90 ) - (-MB 0 ) = MB

M = NIA

W = 200 100 10-6 2.5 10-4 1

= 5 10-6 J = 5

Answer: 5

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Ques. 2JEE Main 2024, 4 April (Evening Shift)

The magnetic moment of a bar magnet is 0.5 Am2. It is suspended in a uniform magnetic field of 8 10-2 T. The work done in rotating it from its most stable to most unstable position is:

4 10-2 J
16 10-2 J
8 10-2 J
Zero

Ques. 3JEE Main 2024, 29 January (Morning Shift)

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 cm from its center is 1.5 10-5 T m. The magnetic moment of the dipole is A m2. (Given : 04 =10-7 T m A-1)

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