JEE MAIN - Physics (2021 - 31st August Morning Shift - No. 5)
A coil having N turns is wound tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. Find the magnetic field at centre, when a current I passes through coil:
$${{{\mu _0}IN} \over {2(b - a)}}{\log _e}\left( {{b \over a}} \right)$$
$${{{\mu _0}I} \over 8}\left[ {{{a + b} \over {a - b}}} \right]$$
$${{{\mu _0}I} \over {4(a - b)}}\left[ {{1 \over a} - {1 \over b}} \right]$$
$${{{\mu _0}I} \over 8}\left( {{{a - b} \over {a + b}}} \right)$$
Explanation
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No. of turns in dx width = $${N \over {b - a}}dx$$
$$\int {dB = \int\limits_a^b {\left( {{N \over {b - a}}} \right)dx{{{\mu _0}I} \over {2x}}} } $$
$$B = {{N{\mu _0}i} \over {2(b - a)}}\ln \left( {{b \over a}} \right)$$
Option (a)
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