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A long straight wire carrying current ...

A long straight wire carrying current of `30 A` is placed in an external unifrom magnetic field of induction `4xx10^(4) T`. The magnetic field is acting parallel to the direction of current. The maggnetic of the resultant magnetic inuduction in tesla at a point `2.0 cm` away form the wire is

A

`10^(-4)`

B

`3xx10^(-4)`

C

`5xx10^(-4)`

D

`6xx10^(-4)`

Text Solution

Verified by Experts

The correct Answer is:
C
Magnetic field due to wire
`B=(mu_(0)I)/(2pi r) = (4pi xx 10^(-7))/(2pi) xx (30)/(2xx10^(-2))`
`=3xx10^(-4)T`
This magnetic field will be perpendicular to external magnetic field.
`:.` Net magnetic field
`B= sqrt(B^(2) +B_(0)^(2))`
`=sqrt((3xx10^(-4))^(2) +(4xx10^(-4))^(2))`
`=5xx10^(-4)T`
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