A long thin walled pipe of radius `R` carries a current I along its length The current density is uniform over the circumference of the pipe The magnetic field at the center of the pipe due to quarter portion of the pipe shown, is
.

Options

A. (a)`(mu_(0)sqrt(2))/(4pi^(2)R)`
B. (b)`(mu_(0)I)/(pi^(2)R)`
C. (c)`(2mu_(0)Isqrt(2))/(pi^(2)R)`
D. (d)None of these

Answer

A

Explanation

`dB cos theta=(mu_(0)(i/(2pir)dx))/(2piR)cos theta , dB sin theta=(mu_(0)i)/(2piR^(2))(dx sin theta)/(2pi)`
`B_(x)=intdBcos theta=(mu_(0)i)/(4pi^(2)R^(2))int_(0)^(pi//2)cos thetaR d theta=(mu_(0)i)/(4pi^(2)R)[sin pi//2-0]=(mu_(0)i)/(4pi^(2)R)`
`B_(Y)=int dB sin theta=int_(0)^(pi/2)(mu_(0)i)/(4pi^(2)R)sin theta d theta =(mu_(0)i)/(4pi^(2)R): B_("Net")=(mu_(0)isqrt(2))/(4pi^(2)R)`