Two coils P and Q each of radius R carry currents I and 8 I respectively in same direction. Those coils are lying in perpendicular planes such that they have a common centre. The magnitude of the ma

Question

Two coils P and Q each of radius R carry currents I and 8 ​ I respectively in same direction. Those coils are lying in perpendicular planes such that they have a common centre. The magnitude of the magnetic field at the common centre of the two coils is ( μ 0 ​ = permeability of free space)

TestHub · Accepted Answer

The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is $B = \frac{μ _{0} I}{2 R}$ .

For coil $P$ , the magnetic field at the centre is $B_{P} = \frac{μ _{0} I}{2 R}$ .

For coil $Q$ , the magnetic field at the centre is $B_{Q} = \frac{μ _{0} ( 8 I )}{2 R} = 8 (\frac{μ _{0} I}{2 R})$ .

Since the coils are in perpendicular planes, their magnetic fields at the common centre are perpendicular to each other.

The magnitude of the resultant magnetic field is $B = B_{P}^{2} + B_{Q}^{2}$ .

$B = (\frac{μ _{0} I}{2 R})^{2} + (8 \frac{μ _{0} I}{2 R})^{2}$

$B = \frac{μ _{0} I}{2 R} 1 + 8 = \frac{3 μ _{0} I}{2 R}$ .

Answer: $\frac{3 μ _{0} I}{2 R}$

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