question_answer

                An electric current is flowing though a circular coil of radius R. The ratio of the magnetic field at the centre of the centre of the coil and that at a distance \[2\sqrt{2}\]R from the centre the centre of the coil and its axis is:                                  JEE Main Online Paper (Held On 09 April 2013)

A)                 \[2\sqrt{2}\]                                                      

B)                 27                                          

C)                 36                                                          

D)                 8

Correct Answer: B

Solution :

                Magnetic at centre \[O=\frac{{{\mu }_{0}}ni}{2R}\] ?(i)                 Magnetic field at point,                 \[P=\frac{{{\mu }_{0}}ni{{R}^{2}}}{2{{[{{R}^{2}}+{{(2\sqrt{2}R)}^{2}}]}^{3/2}}}\]                 ?(ii)                 Dividing Eq. (i) by (ii), we get                 \[=\frac{{{\mu }_{0}}ni}{2R}\times \frac{2\,({{R}^{2}}+8{{R}^{2}})}{{{\mu }_{0}}ni{{R}^{2}}}=\frac{{{(9{{R}^{2}})}^{3/2}}}{{{R}^{3}}}\]                 \[={{(3)}^{3}}=27\]