A short bar magnet with its north pole facing north forms a neutral point at P in the horizontal plane. If the magnet is rotated by `90^(@) ` in the horizontal plane, the net magnetic induction at P is (horizontal component of earth's magnetic field = `B_(H)`)

Options

A. Zero
B. 2 B_(H)
C. `sqrt5/2B_(H)`
D. `sqrt5 B_(H)`

Answer

D

Explanation

When the point P lies on the equatorial line of the magnet
`B_(1)=(mu_(0))/(4pi)(m)/(d^(3))=B_(H)`
When the magnet is rotated by ` 90^(@)`, the magnetic induction at P which is on axial line
`B_(2)=(mu_(0))/(4pi)(2m)/(d^(3))=2B_(H)`
Resultant magnetic induction,
`B=sqrt(B_(1)^(2)+B_(2)^(2))=sqrt((B_(H))^(2)+(2B_(H))^(2))=sqrt5" "B_(H)`