JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 11)
A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear
dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per
unit length of the frame the same, then the self inductance of the coil:
decreases by a factor of $$9\sqrt 3 $$
increases by a factor of 27
decreases by a factor of 9
increases by a factor of 3
Explanation
Total length L will remain constant
L = (3a) N (N = total turns)
and length of winding = (d) N
(d = diameter of wire)
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self inductance = $$\mu $$0n2A$$\ell $$
= $$\mu $$0n2$$\left( {{{\sqrt 3 {a^2}} \over 4}} \right)$$ dN
$$ \propto $$ a2 N $$ \propto $$ a
So self inductance will become 3 times
L = (3a) N (N = total turns)
and length of winding = (d) N
(d = diameter of wire)
self inductance = $$\mu $$0n2A$$\ell $$
= $$\mu $$0n2$$\left( {{{\sqrt 3 {a^2}} \over 4}} \right)$$ dN
$$ \propto $$ a2 N $$ \propto $$ a
So self inductance will become 3 times
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