NEET Physics Magnetism and Matter Questions Solved

NEET - 2009

A bar magnet having a magnetic moment of $2 \times 10^{4} J T^{- 1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B = 6 \times 10^{- 4} T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60 °$ from the field is

(a) 0.6 J (b) 12 J

Concept Videos :-

#3 | Magnetic Dipole Moment & Field due to Bar Magnet

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#4 | Superposition of Magnetic Field due to Bar Magnet

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#5 | Field due to Short Bar Magnet

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#6 | Torque Acting on Bar Magnet in Uniform Field

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#7 | Work Done in Rotating Dipole in Field & PE

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#8 | Oscillations of Bar Magnet in Uniform Field

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#9 | Solenoid as Equivalent Bar Magnet

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Concept Questions :-

Bar magnet

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The work done in rotating a magnetic dipole against the torque acting on it, when placed in magnetic field is stored inside it in the form of potential energy.

When magnetic dipole is rotatd from initial position $θ = θ_{1}$ to final position $θ = θ_{2},$ then work done =MB $(\cos θ_{1} - \cos θ_{2})$

=MB $(1 - \frac{1}{2})$

= $\frac{2 \times 10^{4} \times 6 \times 10^{- 4}}{2} = 6 J$

Difficulty Level:

(a)
13%
(b)
14%
(c)
72%
(d)
3%

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