1.

A long wire carries a steady curent . It is bent into a circle of one turn and the magnetic field at the centre of the coil is `B`. It is then bent into a circular loop of `n` turns. The magnetic field at the centre of the coil will beA. n BB. `n^2B`C. `2 n B`D. `2 n^2 B`

Answer» Correct Answer - B
For same length , `B prop n^2`
`therefore (B_2)/(B_1)=((n_2)/(n_1))^2`
`B_2=((n)/(1))^2xxB=n^2B`.


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