A current I is flowing in a straight conductor of length L. The magnetic induction at a point distant `(L)/(4)` from its centre will be

Explanation

We have the result of magnetic filed for a straight conductor. For a distribution, we have to make a differential equation and integrate.
Consider a small element of width `R d theta` as shown in the figure.
`dB=(mu_(0)(dI))/(2piR)`
`dI=(I)/((piR)/(2))d theta=(2I)/(pi)d theta :. dB=(mu_(0)Id theta)/(pi^(2)R)`
As we have to integrate `x` and `y` components separately.
`dB_(x)=dB costheta=(mu_(0)I)/(pi^(2)R)costheta d theta`
`dB_(y)=dB sin theta=(mu_(0)I)/(pi^(2)R)sintheta d theta`
`B_(x)=intdB_(x)=(mu_(0)I)/(pi^(2)R)int_(-pi//4)^(+pi//4)costheta d theta=(mu_(0)Isqrt(2))/(pi^(2)R)`
`B_(y)=intdB_(y)=(mu_(0)I)/(pi^(2)R)int_(-pi//4)^(+pi//4)sintheta d theta=0`
`vecB=(mu_(0)Isqrt(2))/(pi^(2)R)hati`