There is one circular loop of wire of radius R, carrying current I and magnetic field at its centre is `B_(0)`, due to its own current. The loop is bent along its diameter at right angle. What will be the net magnetic field at the centre of the loop?

Options

A. `B_(0)/sqrt2`
B. `B_(0) sqrt2`
C. `2B_(0)sqrt2`
D. `B_(0)sqrt3`

Answer

a

Explanation

Magnetic field at the centre of the semicircular loop will be `B_(0)//2`, half of that at the centre of the circular loop. When one semi-circular part of the loop is turned by `90^(@)`, the magnetic fields due to two semi-circular parts are acting at the centre in mutually perpendicular directions. Hence, the net magnetic field will be a resultant of the two:
`B = B_(0)/2 sqrt2 = B_(0)/sqrt2`
Hence, option (a) is correct.