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Question

(C) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_{0} I}{2r}$.Given $B_{P} = B_{Q}$,we have $\frac{\mu_{0} I_{P}}{2 r

Vedclass · Accepted Answer

$V_{Q} = 4 V_{P}$

Vedclass · Answer

(C) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_{0} I}{2r}$.Given $B_{P} = B_{Q}$,we have $\frac{\mu_{0} I_{P}}{2 r_{P}} = \frac{\mu_{0} I_{Q}}{2 r_{Q}}$.Since $r_{Q} = 2 r_{P}$,we get $\frac{I_{P}}{r_{P}} = \frac{I_{Q}}{2 r_{P}}$,which implies $I_{Q} = 2 I_{P}$.The resistance of a wire is $R = ho \frac{L}{A} = ho \frac{2 \pi r}{A}$. Since the wire is similar,$ho$ and $A$ are constant,so $R \propto r$.Therefore,$R_{Q} = 2 R_{P}$.The potential difference is $V = IR$. Thus,$\frac{V_{Q}}{V_{P}} = \frac{I_{Q} R_{Q}}{I_{P} R_{P}} = \left(\frac{2 I_{P}}{I_{P}}ight) 	imes \left(\frac{2 R_{P}}{R_{P}}ight) = 2 	imes 2 = 4$.Hence,$V_{Q} = 4 V_{P}$.

Two circular coils $P$ and $Q$ are made from similar wire,but the radius of $Q$ is twice that of $P$. What should be the value of the potential difference across them so that the magnetic induction at their centres is the same?

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