Question

A solid cylindrical conductor of radius ${}^{\prime }{R}^{\prime }$ has a uniform current density. The magnetic field ${}^{\prime }{H}^{\prime }$ insider the conductor at a distance ${}^{\prime }{r}^{\prime }$ from the axis of the conductor is:

• A. $I/2\pi r$
• B. $I/4\pi r$
• C. $Ir/2\pi R^2$
• D. $Ir/4\pi R^2$

Answer 1

The correct option is C $Ir/2\pi R^2$

Magnetic field at a distance $r$ inside a conductor of radius $R$ is given by ampere’s circuital law

$B\times 2\pi r={\mu }_{0}J\pi r^2$

$B=\frac{{\mu }_{0}Jr}{2}$

$B=\frac{{\mu }_{0}rI}{2\pi R^2}$

Thus magnetic field inside a conductor is given by

$B=\frac{{\mu }_{0}rI}{2\pi R^2}$

Magnetic field strength

$H=\frac{B}{{\mu }_0}$

$H=\frac{Ir}{2\pi R^2}$