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Question

If the magnetic field at ‘P’ can be written as K tan(α2), then K is

  1. μ0I4πd
  2. μ0I2πd
  3. μ0Iπd
  4. 2μ0I4πd


Solution

The correct option is B μ0I2πd
Magnetic field at point P will be due to two elements AC and AB which will be in same direction as out of the plane as shown in diagram. 

We know that magnetic field due to current carrying wire is 
B=μ0I4πd(sinθ1+sinθ2)

So for wire AC, θ1=(90α)
θ2=90
BAC=μ0I4πdsinα(sin((90α)+sin90)

BAC=μ0I4πdsinα(1cosα)
BAC=μ0I4πd(tanα2)

Now net magnetic field at point P will be double of the magnetic field due to element AC due to symmetry. 
BP=μ0I2πd(tanα2)

K=μ0I2πd
 

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