Question

The magnetic field due to a long current carrying wire at a perpendicular distance $R$ is $B,$ then the magnetic field at perpendicular distance $2R$ will be equal to

$[$0.77 Mark$]$

• A. $2B$
• B. $B$
• C. $\frac{B}{4}$
• D. $\frac{B}{2}$

Answer 1

The correct option is D $\frac{B}{2}$

Magnetic field due to a long current carrying wire at a perpendicular distance $R$ is given by
$B=\frac{{\mu }_{0}i}{2\pi R}$
At distance $2R,$
$B^{\prime }=\frac{{\mu }_{0}i}{2\pi \left(2R\right)}$
$\therefore B^{\prime }=\frac{B}{2}$