Two identical coils P and Q each of radius R are lying in perpendicular planes such that they have common centre. Find the magnitude and direction of magnetic field at the common centre of the two coils, if they carry currents equal to I and square root of 3I respectively.

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Dear Student,
Answer to your question is given below:
$G i v e n t h a t t w o i d e n t i c a l c o i l s a r e l y i n g i n p e r p e n d i c u l a r p l a n e s a n d h a v i n g c o m m o n c e n t r e . P a n d Q c a r r y c u r r e n t I a n d \sqrt{3 I} r e s p e c t i v e l y . N o w, m a g n e t i c f i e l d a t t h e c e n t r e o f P d u e t o i t s c u r r e n t I i s, B_{P} = \frac{μ_{o} I}{2 R} A n d, m a g n e t i c f i e l d a t c e n t r e o f Q, d u e t o i t s c u r r e n t \sqrt{3 I} i s, B_{Q} = \frac{μ_{o} \sqrt{3 I}}{2 R} N o w t h e n e t m a g n e t i c f i e l d i s, {(B_{n e t})}^{2} = {(B_{P})}^{2} + {(B_{Q})}^{2} \Rightarrow B_{n e t} = \sqrt{{(\frac{μ_{o} I}{2 R})}^{2} + {(\frac{μ_{o} \sqrt{3 I}}{2 R})}^{2}} \Rightarrow B_{n e t} = \sqrt{{(\frac{μ_{o} I}{2 R})}^{2} 4} \Rightarrow B_{n e t} = \frac{μ_{o} I}{R} N o w t h e d i r e c t i o n o f t h e n e t m a g n e t i c f i e l d i s, \tan (θ) = \frac{B_{P}}{B_{Q}} \Rightarrow \tan (θ) = \frac{\frac{μ_{o} I}{2 R}}{\frac{μ_{o} \sqrt{3 I}}{2 R}} \Rightarrow \tan (θ) = \frac{1}{\sqrt{3}} \Rightarrow θ = \tan^{- 1} (\frac{1}{\sqrt{3}}) \Rightarrow θ = 30^{o}$
Regards,

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