A long, straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of `4.0 xx 10^-4` T exists from south to north.Find at a point 2.0 away from the wire.

Options

A. `10^(-4)`
B. `3xx10^(-4)`
C. `5xx10^(-4)`
D. `6xx10^(-4)`

Answer

C

Explanation

Given `I=30 A, B_(1)=4xx10^(-4)T`
`R=2 cm =0.02 m`
Magnetic field strength at a point `P` due to current carrying wire,
`B_(2)=(mu_(0)I)/(2pir)=(4pixx10^(-7)xx30)/(2pixx0.02)=3xx10^(-4)T`
The direction of `B_(2)` will be perpendicular to `B_(1)`, hence, resulatant field at

`P=sqrt(B_(1)^(2)+B_(2)^(2))`
`=sqrt(4^(2)+3^(2))xx10^(-4)=5xx10^(-4)T`