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  • A bar magnet with poles 25cm apart and of strength 14.4 amp-m rests with the centre on a frictionless pivot. It is held in equilibrium at an angle of 30° with respect to a uniform magnetic fiel

A bar magnet with poles 25cm apart and of strength 14.4 amp-m rests with the centre on a frictionless pivot. It is held in equilibrium at an angle of 30° with respect to a uniform magnetic field of induction 0.25 Wb/m2, by applying a force F at right angles to its axis at a point 12cm from pivot. Find the force.

Option: 1

25N


Option: 2

15.90N


Option: 3

15N


Option: 4

30N


Answers (1)

best_answer

Here, 2l = 25 cm 

\implies l = \frac{25}{2} cm = \frac{25}{200}m

Also, m * 2l = 14.4

\implies m = \frac{14.4}{0.25} = 57.6 A-m^{2}

Force is the torque due to the magnetic field 

                = P_{m} * B * \sin 30\degree = F * 0.12

So,

F * 0.12 = 14.4 * 0.2 J * \sin 30\degree

or

F =\frac{14.4 * 0.25 * \frac{1}{2}}{0.12} = 15 N

Posted by

Gautam harsolia

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