Two identical concentric coils $X$ and $Y$ carrying currents in the ratio $1 : 2$ are arranged in mutually perpendicular planes. If the magnetic field due to coil $X$ is $B$ the net field at their common centre is:

B

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2 B

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3 B

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\sqrt{5} B

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Solution

The correct option is B $\sqrt{5} B$
As the coil are arranged in mutually perpendicular planes, the field produced at their common centre $B_{X}$ and $B_{Y}$ will be perpendicular to each other.
More over, except current, all other parameters being same.

$\frac{B_{X}}{B_{Y}} = \frac{I_{X}}{I_{Y}} = \frac{1}{2}$

$\Rightarrow B_{X} = B$ & $B_{Y} = 2 B$
The net magnetic field is given by, $B_{n e t} = \sqrt{B_{X}^{2} + B_{Y}^{2}} = \sqrt{B^{2} + {(2 B)}^{2}} = \sqrt{5} B$

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Two identical concentric coils X and Y carrying currents in the ratio 1:2 are arranged in mutually perpendicular planes. If the magnetic field due to coil X is B the net field at their common centre is:

Two identical concentric coils $X$ and $Y$ carrying currents in the ratio $1 : 2$ are arranged in mutually perpendicular planes. If the magnetic field due to coil $X$ is $B$ the net field at their common centre is: