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Question

# Three circular coils each having two turns and radius R are placed mutually perpendicular to each other and each having its centre at the origin of the coordinate system. If current i is flowing through each coil, then the magnitude of the magnetic field at the common centre is:

A
3μ0iR
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B
Zero
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C
21μ0i2R
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D
(32)μ0i2R
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Solution

## The correct option is A √3μ0iRThe magnetic field due to a circular current carrying coil of n turns at its centre is, B=μ0ni2R Hence, due to each coil the magnitude of magnetic field at centre O will be B=2μ0i2R=μ0iR (∵n=2) Now irrespective of the direction of current in coils, the direction of magnetic field will be mutually perpendicular due to each coil. Using right-hand thumb rule and the assumed direction of current as shown in figure. Magnetic field due to coil 1, coil 2 & coil 3 will be: B1=B^k; B2=B^i; B3=B^j Thus, the net magnetic field at centre O will be →Bnet=→B1+→B2+→B3 →Bnet=B^k+B^i+B^j |→Bnet|=√B2+B2+B2=√3B After putting the value of B, we get ∴|→Bnet|=√3μ0iR Hence, option (a) is correct.

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