Currents through ABC and `A^'B^'C^'` is I, figure. What is the magnetic field at P? `BP=PB^'=r`. (Here `C^'B^'PBC` are collinear)

Options

A. `(2I)/(4pir)`
B. `(mu_0)/(4pi)((2I)/(r))`
C. `(mu_0)/(4pi)(1/r)`
D. zero

Answer

B

Explanation

The magnetic field induction at P due to current I in AB is `B_1=(mu_0)/(4pi)I/r`
Acting perpendicular to the plane of paper downwards.
Magnetic field induction at P due to current I in BC is, `B_2=0` (as point P lies on the line of BC)
Magnetic field inductin at P due to current I in `A^'B^'` is

`B_3=(mu_0)/(4pi)I/r` (acting perpendicular to the plane of paper downwards)
Magnetic field induction at P due to current I in `B^'C^'` is `B_4=0` (as point P lies on the line of `B^'C^'`)
Thus total magnetic field induction at P is
`B=B_1+B_2+B_3+B_4`
`=(mu_0)/(4pi)I/r+0+(mu_0)/(4pi)I/r+0=(mu_0)/(4pi)(2I)/(r)`