A straight wire current element is carrying current `100A`, as shown in figure. The magnitude of magnetic field at point `P` which is at perpendicular distance `(sqrt(3)-1)m` from the current element if end `A` and end `B` of the element subtend angle `30^(@)` and `60^(@)` at point `P`, as shown, is:

Options

A. `5xx10^(-6) T`
B. `2.5xx10^(-6) T`
C. `2.5xx10^(-5) T`
D. `8xx10^(5) T`

Answer

A

Explanation

`B=(mu_(0).i)/(4pid) (sin theta_(1)+sin theta_(2))`
`=(10^(-7)xx100)/(sqrt(3)-1)[(sqrt(3))/2-1/2]=5xx10^(-6)T`