question_answer

An electric current is flowing through 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 ONLINE 09-04-2013]

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

B) \[27\]

C)             \[36\]

D) \[8\]

Correct Answer: B

Solution :

[b] 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

\[=\frac{{{\mu }_{0}}ni}{2R}\times \frac{2\,({{R}^{2}}+8{{R}^{2}})}{{{\mu }_{0}}ni{{R}^{2}}}=\frac{{{(9{{R}^{2}})}^{3/2}}}{{{R}^{3}}}\]

\[={{(3)}^{3}}=27\]