A circular coil of radius $R$ carries a current $i$ . The magnetic field at its centre is $B$ . The distance from the centre on the axis of the coil where the magnetic field will be $B / 8$ is

\sqrt{2} R

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\sqrt{3} R

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2 R

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3 R

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Solution

The correct option is D $\sqrt{3} R$
$B = \frac{2 μ_{0} i \cdot π r^{2}}{4 π (r^{2} + x^{2}) \frac{3}{2}}$
magnetic field at centre $= \frac{μ_{0} i}{2 r}$
$B = \frac{μ_{0} i}{2 r}$
Put $r = \sqrt{3} R$
$B^{1} = \frac{2 μ_{0} i π r 2}{4 π (4 r^{2}) \frac{3}{2}}$
$= \frac{B}{8}$

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A circular coil of radius R carries a current i. The magnetic field at its centre is B. The distance from the centre on the axis of the coil where the magnetic field will be B/8 is

The correct option is D √3RB=2μ0i⋅πr24π(r2+x2)32magnetic field at centre =μ0i2rB=μ0i2rPut r=√3RB1=2μ0iπr24π(4r2)32=B8

A circular coil of radius $R$ carries a current $i$ . The magnetic field at its centre is $B$ . The distance from the centre on the axis of the coil where the magnetic field will be $B / 8$ is

The correct option is D $\sqrt{3} R$
$B = \frac{2 μ_{0} i \cdot π r^{2}}{4 π (r^{2} + x^{2}) \frac{3}{2}}$
magnetic field at centre $= \frac{μ_{0} i}{2 r}$
$B = \frac{μ_{0} i}{2 r}$
Put $r = \sqrt{3} R$
$B^{1} = \frac{2 μ_{0} i π r 2}{4 π (4 r^{2}) \frac{3}{2}}$
$= \frac{B}{8}$