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Standard XII

Physics

Magnetic Field Due to a Straight Long Wire

The magnetic ...

Question

The magnetic field at the center of a circular coil of radius r carrying current is $B_{1}$ . The field at the center of another coil of radius $\frac{r}{2}$ carrying same current is $B_{2}$ then ratio of $B_{1} / B_{2}$ is :-

1:2

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2:1

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1:1

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8:1

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Solution

The correct option is D 8:1
A magnetic field is inversely proportional to the r³ of the radius of the coil

So $B 1 / B 2 = (1 / r^{3}) / (1 / (r / 2)^{3})$

$B 1 / B 2 = 8 / 1$

$B 1 : B 2 = 8 : 1$

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The magnetic field at the center of a circular coil of radius r carrying current is B1. The field at the center of another coil of radius r2 carrying same current is B2 then ratio of B1/B2 is :-

The correct option is D 8:1A magnetic field is inversely proportional to the r3 of the radius of the coil So B1/B2=(1/r3)/(1/(r/2)3) B1/B2=8/1 B1:B2=8:1

The magnetic field at the center of a circular coil of radius r carrying current is $B_{1}$ . The field at the center of another coil of radius $\frac{r}{2}$ carrying same current is $B_{2}$ then ratio of $B_{1} / B_{2}$ is :-

The correct option is D 8:1
A magnetic field is inversely proportional to the r³ of the radius of the coil

So $B 1 / B 2 = (1 / r^{3}) / (1 / (r / 2)^{3})$

$B 1 / B 2 = 8 / 1$

$B 1 : B 2 = 8 : 1$