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Question

Let $r_{1}$ and $r_{2}$ are the radius of coil $1$ and $2$. If $B_{1}$ and $B_{2}$ are magnetic induction at their centre, then

$B_{1}=\frac{\mu_{0

Vedclass · Accepted Answer

$4$

Vedclass · Answer

Let $r_{1}$ and $r_{2}$ are the radius of coil $1$ and $2$. If $B_{1}$ and $B_{2}$ are magnetic induction at their centre, then

$B_{1}=\frac{\mu_{0} I_{1}}{2 r_{1}} ;$ and $B_{2}=\frac{\mu_{0} I_{2}}{2 r_{2}}$

Since $B_{1}=B_{2} ;$ and $r_{1}=2 r_{2}$

$I_{1}=2 I_{2}$

Again if $R_{1}$ and $R_{2}$ are resistance of the coil $1$ and $2$ then $R_{1}=2 R_{2}($ as $R \propto$ length $=2 \pi r)$ and if $V_{1}$ and $V_{2}$ are the potential difference across them respectively, then

$\frac{V_{1}}{V_{2}}=\frac{I_{1} R_{1}}{I_{2} R_{2}}=\frac{\left(2 I_{2}\right)\left(2 R_{2}\right)}{I_{2} R_{2}}=4$

Two circular coils $1$ and $2$ are made from the same wire but the radius of the $1^{st}$ coil is twice that of the $2^{nd}$ coil. What is the ratio of potential difference in volts should be applied across them so that the magnetic field at their centres is the same?

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