Two wires $A\ and\ B$ are of lengths 40 cm and 30 cm. $A$ is bent into a circle of radius $r$ and $B$ into an arc of radius $r.$ A current $i_{1}$ is passed through $A$ and $i_{2}$ through $B.$ To have the same magnetic inductions at the centre, the ratio of $i_{1}\ :i_{2}$ is

Question

Dr. Abubacker · Accepted Answer

(a) For wire $A,$
$B_{1} = \frac{\mu_{0}i_{1}}{2r}$
Where $r = \frac{40}{2\pi}$
For wire $B,$
Circumference = length
$\ \ \ \ \ \ n\pi r = 30$
or $\ \ n\pi = \frac{30}{r} = \frac{30}{40/2\pi} = \frac{3}{2}\pi$
or $\ \ \ 	heta = n\pi = \frac{3}{2}\pi$
$	herefore\ \ \ \ \ B_{2} = \frac{\mu_{0}}{4\pi}\left( \frac{i_{2}}{r} ight)	heta$
But ${\ B}_{1} = B_{2}$
or $\frac{\mu_{0}i_{1}}{2r} = \ \frac{\mu_{0}}{4\pi}\left( \frac{i_{2}}{r} ight)	heta$
$or\ \ \ \frac{i_{1}}{i_{2}} = \frac{3}{4}$

Q. Two wires $A\ and\ B$ are of lengths 40 cm and 30 cm. $A$ is bent into a circle of radius $r$ and $B$ into an arc of radius $r.$ A current $i_{1}$ is passed through $A$ and $i_{2}$ through $B.$ To have the same magnetic inductions at the centre, the ratio of $i_{1}\ :i_{2}$ is

Explanation

Reviewed by Dr. Abubacker

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