A straight wire carrying a current `10 A` is bent into a semicircular arc of radius `5 cm`. The magnitude of magnetic field at the center is

Options

A. `8A `
B. `4A`
C. `2A`
D. `1A`

Answer

A

Explanation

`theta =(arc)/("radius")=(0.1)/(0.2)=1/2` radian
`therefore` The dield (B) at the centre of a circular coil corresponds to `2pi` radian s.
`therefore F or 1/2` radian, the field will be `B' (B)/(2xx2pi) =(B)/(4pi)`
Let I be teh current to produce the field,
`B' =2xx 10^(-6) T`
`therefore B ' =(1)/(4pi) [(mu_(0))/(4pi) (2piI)/(r)]`
`therefore 2xx10^(-6) =(1)/(4pi) xx10^(-7) xx(2 piI)/(0.2) =(10^(-7) I)/(2xx0.2)`
`therefore I =(2xx2xx0.2xx10^(-6))/(10^(-7))=8A`
Note: The magnetic field dut to an are substending an angle `theta` at the centre is `((theta)/(2pi) B).`
In this `cose theta =1/2` raian and `B ' =(B)/(4pi).`