question_answer

(i) A circular coil of 30 turns and radius 8 cm carrying a current of 6 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1 T. The field lines make an angle \[60{}^\circ \] with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

(ii) Would your answer change, if the circular coil were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered).

Or

Explain the following.

(i) Why do magnetic lines form continuous closed loops?

(ii) Why are the field lines repelled (expelled) when a diamagnetic materials is placed in an external uniform magnetic field?

Answer:

            Here,     N = 30, \[R=8\,cm=8\times {{10}^{-2}}m,\] I = 6 A, \[\theta =60{}^\circ \] and B = 1 T (i) The magnitude of the counter torque = magnitude of the deflecting torque \[=NAIB\,\,\sin \theta =N\cdot (\pi {{R}^{2}})IB\sin \theta \]             \[=30\times 3.14\times {{(8\times {{10}^{-2}})}^{2}}\times 6\times 1\times \sin 60{}^\circ \]             \[=3.13\,N\text{-}m\] (ii) The answer would not change as area enclosed by the coil as well as all other particular remain unaltered and the formula, \[\tau =\text{ }NAIB\,sin\,\theta \]is true for planar coil for any shape. Or (i) Magnetic lines comes out from North pole and enter into the South pole outside the magnet and travel from South pole to North pole inside the magnet. Hence, magnetic lines forms closed loops. (ii) The diamagnetic material gets slightly magnetised in a direction opposite to external field, therefore magnetic lines are repelled by diamagnetic material.