Question

A long straight wire carrying a current of $30A$ is placed in an external uniform magnetic field of induction $4\times 10^{-4}T$. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point $2.0cm$ away from the wire is $\left({\mu }_0=4\pi \times 10^{-7}H{m}^{-1}\right)$

• A. ${10}^{-4}$
• B. $3\times {10}^{-4}$
• C. $5\times {10}^{-4}$
• D. $6\times {10}^{-4}$

Answer 1

Answer: (C)

Given, $i=30A,B_1=4\times 10^{-4}T$,
$r=2cm=0.02m$

Magnetic field induction at point $P$ due to current carrying wire is
$B_2=\frac{{\mu }_{0}i}{2\pi r}=\frac{4\pi \times 10^{-7}\times 30}{2\pi \times 0.02}$
$=3\times 10^{-4}T$
The direction of $B_2$ will be perpendicular to $B_1$, then the magnitude of the resultant magnetic induction
$B=\sqrt{B_1^2+B_2^2}$
$=\left(\sqrt{(4{)}^2+(3{)}^2}\right)\times 10^{-4}$
$B=5\times 10^{-4}T$