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#4 | Superposition of Magnetic Field due to Bar Magnet

(Physics) > Magnetism and Matter

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The magnetic field at a point x on the axis of a small bar magnet is equal to the field at a point y on the equator of the same magnet. The ratio of the distances of x and y from the centre of the magnet is

(a) ${2}^{-3}$

(b) ${2}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

(c) ${2}^{3}$

(d) ${2}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

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Points A and B are situated perpendicular to the axis of a 2cm long bar magnet at large distances X and 3X from its centre on opposite sides. The ratio of the magnetic fields at A and B will be approximately equal to

(a) 1 : 9 (b) 2 : 9

(c) 27 : 1 (d) 9 : 1

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Two short magnets with their axes horizontal to the magnetic meridian are placed with their centres 40 *cm *east and 50 *cm* west of magnetic needle. If the needle remains undeflected, the ratio of their magnetic moments is

(a) 4:5

(b) 16:25

(c) 64:125

(D) 2:$\sqrt{5}$

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If a bar magnet of magnetic moment *M* is freely suspended in a uniform magnetic field of strength *B*, the work done in rotating the magnet through an angle is

(a) $MB(1-\mathrm{sin}\theta )$

(b) $MB\mathrm{sin}\theta $

(c) $MB\mathrm{cos}\theta $

(d) $MB(1-\mathrm{cos}\theta )$

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