B_{X} and B_{Y} are the magnetic fields at th | vedclass

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

Vedclass · Accepted Answer

$1: 2$

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(B) The magnetic field at the centre of a circular coil with $N$ turns,radius $R$,and current $i$ is given by the formula: $B = N \left( \frac{\mu_{0} i}{2R} ight)$.For coil $X$: $N_{X} = 200$,$R_{X} = 20 \ cm$,current $= i$. Thus,$B_{X} = 200 \left( \frac{\mu_{0} i}{2 	imes 20 \ cm} ight)$.For coil $Y$: $N_{Y} = 400$,$R_{Y} = 20 \ cm$,current $= i$. Thus,$B_{Y} = 400 \left( \frac{\mu_{0} i}{2 	imes 20 \ cm} ight)$.Taking the ratio of $B_{X}$ to $B_{Y}$:$\frac{B_{X}}{B_{Y}} = \frac{200 \left( \frac{\mu_{0} i}{2 	imes 20 \ cm} ight)}{400 \left( \frac{\mu_{0} i}{2 	imes 20 \ cm} ight)} = \frac{200}{400} = \frac{1}{2}$.Therefore,the ratio is $1: 2$.

$B_{X}$ and $B_{Y}$ are the magnetic fields at the centre of two coils $X$ and $Y$ respectively,each carrying equal current. If coil $X$ has $200$ turns and $20 \ cm$ radius and coil $Y$ has $400$ turns and $20 \ cm$ radius,the ratio of $B_{X}$ and $B_{Y}$ is:

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