AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

Consider the figure shown below,The magnetic field at point P due to Ix is,
Bx=−

µ04

2πd

^

j

Here, the negative sign is taken as the direction of magnetic field is in −y direction.
The magnetic field at point P due to Iy is,
By=

µ03

2πd

^

i

The resultant magnetic field is,
B=√Bx2+By2
Substitute the value sin above expression.
B=√(−

µ04

2πd

)2+(

µ03

2πd

)2
B=

µ0

2πd

√42+32
B=

µ05

2πd