JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 11)
One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the radio of the magnetic field at the central of the loop (BL) to that at the center of the coil (BC), i.e. $${{{B_L}} \over {{B_C}}}$$ will be :
N
$${1 \over N}$$
N2
$${1 \over {{N^2}}}$$
Explanation
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For loop,
L = 2$$\pi $$R
For coil,
L = N $$ \times $$ 2$$\pi $$r
$$ \therefore $$ 2$$\pi $$R = N $$ \times $$ 2$$\pi $$r
$$ \Rightarrow $$ R = Nr
$$ \Rightarrow $$ r = $${R \over N}$$
We know,
BL = $${{{\mu _0}i} \over {2R}}$$
and BC = N $$ \times $$ $${{{\mu _0}i} \over {2r}}$$
$$ \therefore $$ $${{{B_L}} \over {{B_C}}} = {{{{{\mu _0}i} \over {2R}}} \over {N \times {{{\mu _0}i} \over {2\left( {{R \over N}} \right)}}}} = {1 \over {{N^2}}}$$
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