Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2 I, respectively. The resultant magnetic field induction at the centre will be

\frac{\sqrt{5} μ_{0} I}{2 R}

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\frac{3 μ_{0} I}{2 R}

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\frac{μ_{0} I}{2 R}

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\frac{μ_{0} I}{R}

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Solution

The correct option is A $\frac{\sqrt{5} μ_{0} I}{2 R}$
The magnetic field, due the coil, carrying current I Ampere
$B_{1} = \frac{μ_{0} I}{2 R}$
The magnetic field due to the coil, carrying current 2I Ampere
$B_{2} = \frac{μ_{0} (2 I)}{2 R}$
The resultant B
$B_{n e t} = \sqrt{B_{1}^{2} + B_{2}^{2} + 2 B_{1} B_{2} c o s θ}$

where $θ = 90^{o}$
$\Rightarrow B_{n e t} = \sqrt{B_{1}^{2} + B_{2}^{2} + 2 B_{1} B_{2} cos 90^{0}} = \frac{μ_{0} (2 I)}{2 R} \sqrt{1 + 4}$
$= \frac{\sqrt{5} μ_{0} I}{2 R}$

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Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2 I, respectively. The resultant magnetic field induction at the centre will be

The correct option is A √5μ0I2RThe magnetic field, due the coil, carrying current I AmpereB1=μ0I2RThe magnetic field due to the coil, carrying current 2I AmpereB2=μ0(2I)2RThe resultant BBnet=√B21+B22+2B1B2cosθ where θ=90o⇒Bnet=√B21+B22+2B1B2cos900=μ0(2I)2R√1+4=√5μ0I2R