A circular coil connected to a battery of emf | vedclass

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Question

(A) The magnetic field at the centre of a circular coil with $n$ turns is given by $B = \frac{\mu_0 i n}{2 r}$.Let the original length be $L = 2 \pi r

Vedclass · Accepted Answer

$\frac{2 E}{9}$

Vedclass · Answer

(A) The magnetic field at the centre of a circular coil with $n$ turns is given by $B = \frac{\mu_0 i n}{2 r}$.Let the original length be $L = 2 \pi r n$. When the wire is stretched to double its length,$L^{\prime} = 2L = 4 \pi r n$.The new radius is $r^{\prime} = r/3$. The new number of turns $n^{\prime}$ is given by $L^{\prime} = 2 \pi r^{\prime} n^{\prime} \Rightarrow 4 \pi r n = 2 \pi (r/3) n^{\prime} \Rightarrow n^{\prime} = 6n$.For the same magnetic field $B$,$\frac{\mu_0 i n}{2 r} = \frac{\mu_0 i^{\prime} n^{\prime}}{2 r^{\prime}}$.Substituting $n^{\prime} = 6n$ and $r^{\prime} = r/3$,we get $\frac{i n}{r} = \frac{i^{\prime} (6n)}{r/3} \Rightarrow \frac{i n}{r} = \frac{18 i^{\prime} n}{r} \Rightarrow i^{\prime} = \frac{i}{18}$.Since the volume of the wire remains constant,$A L = A^{\prime} L^{\prime} \Rightarrow A L = A^{\prime} (2L) \Rightarrow A^{\prime} = A/2$.The new resistance is $R^{\prime} = \rho \frac{L^{\prime}}{A^{\prime}} = \rho \frac{2L}{A/2} = 4 \rho \frac{L}{A} = 4R$.The new emf is $E^{\prime} = i^{\prime} R^{\prime} = (i/18) (4R) = \frac{2}{9} (iR) = \frac{2}{9} E$.

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