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Question

(B) The magnetic field at the centre of a circular coil with $N$ turns and radius $R$ carrying current $i$ is given by $B = \frac{N \mu_{0} i}{2 R}$.F

Vedclass · Accepted Answer

$\frac{25}{4}$

Vedclass · Answer

(B) The magnetic field at the centre of a circular coil with $N$ turns and radius $R$ carrying current $i$ is given by $B = \frac{N \mu_{0} i}{2 R}$.For the first coil: $B_{1} = \frac{N_{1} \mu_{0} i}{2 R_{1}}$,where $N_{1} = 2$.When the wire is unwound and rewound,the total length of the wire $L = N_{1} (2 \pi R_{1}) = N_{2} (2 \pi R_{2})$ remains constant.Thus,$R_{2} = R_{1} \frac{N_{1}}{N_{2}} = R_{1} \frac{2}{5}$.For the second coil: $B_{2} = \frac{N_{2} \mu_{0} i}{2 R_{2}}$,where $N_{2} = 5$.Taking the ratio: $\frac{B_{2}}{B_{1}} = \frac{N_{2}}{N_{1}} 	imes \frac{R_{1}}{R_{2}} = \frac{N_{2}}{N_{1}} 	imes \frac{N_{2}}{N_{1}} = \left( \frac{N_{2}}{N_{1}} ight)^{2}$.Substituting the values: $\frac{B_{2}}{B_{1}} = \left( \frac{5}{2} ight)^{2} = \frac{25}{4}$.

The electric current in a circular coil of $2$ turns produces a magnetic induction $B_{1}$ at its centre. The coil is unwound and is rewound into a circular coil of $5$ turns and the same current produces a magnetic induction $B_{2}$ at its centre. The ratio of $\frac{B_{2}}{B_{1}}$ is:

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