The work done in moving a dipole from its most stable to most unstable position in a 0.09 T uniform magnetic field is (dipole moment of this dipole =

Question

The work done in moving a dipole from its most stable to most unstable position in a 0.09 T uniform magnetic field is (dipole moment of this dipole =

The work done in moving a dipole from its most stable to most unstable position in a 0.09 T uniform magnetic field is (dipole moment of this dipole = `0.5 Am^(2))`
A. 0.07J
B. 0.08J
C. 0.09J
D. 0.1J

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
Since the most stable position is at `theta=0` and the most unstable position is at `theta=180^(@)`, then the work done is given by
`w=underset(theta=0^(@))overset(theta=180^(@))inttau(theta)d theta=underset(theta=0^(@))overset(theta=180^(@))int mB sin theta d theta=-mB[cos theta]_(0)^(180^(@))`
`=-mB[cos 180^(@)-cos 0^(@)]=-mB[-1-1]`
`=-mB[-2]=2mB`
`therefore W=2xx0.50xx0.09=0.09J`