JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 19)
There are two long co-axial solenoids of same length $$l$$. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2, respectively. The ratio of mutual inductance to the self - inductance of the inner-coil is :
$${{{n_2}} \over {{n_1}}}.{{{r_2}^2} \over {{r_1}^2}}$$
$${{{n_2}} \over {{n_1}}}$$
$${{{n_1}} \over {{n_2}}}$$
$${{{n_2}} \over {{n_1}}}.{{{r_1}} \over {{r_2}}}$$
Explanation
$$M = {\mu _0}\,{n_1}\,{n_2}\,\pi r_1^2$$
$$L = {\mu _0}\,n_1^2\,\pi r_1^2$$
$$ \Rightarrow \,\,{M \over L} = {{{n_2}} \over {{n_1}}}$$
$$L = {\mu _0}\,n_1^2\,\pi r_1^2$$
$$ \Rightarrow \,\,{M \over L} = {{{n_2}} \over {{n_1}}}$$
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