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. 0.07JB. 0.08JC. 0.09JD. 0.1J

The work done in moving a dipole from its most stable to most unstable

1.	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. 0.07JB. 0.08JC. 0.09JD. 0.1J
Answer» 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`

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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. 0.07JB. 0.08JC. 0.09JD. 0.1J

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