AP EAPCET 08-Jul-2022 Shift 1 Paper

The situation given can be shown as

Magnetic field at some point P with position vector r due to a straight long current carrying wire is given by
B=

µ0I

2πd

Where, d= Perpendicular distance of point from wire.
Here, r=2

∧

i

+4

∧

j

m
⇒ for wire on X - axis, d=dx=4m
and for wire on Y-axis d=dy=2m
As magnetic fields due to wires along X and along Y-axis are in opposite direction ( RH rule) so,
Bnet=B+Bx
=

µ0I2

2πdy

(

∧

k

)+

µ0I1

2πdx

(−

∧

k

)
=

µ0

2π

(

I2

dy

−

I1

dx

)

∧

k

=

4π×10−7

2π

(

6

2

−

8

4

)

∧

k

=2×10−7

∧

k

T
∴ |Bnet|=2×10−7T