Question

A long straight wire carrying a current of $30\, A$ 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 $2.0\, cm$ 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, $i=30\, A, B_{1}=4 \times 10^{-4} T$,
$r=2\, c m=0.02\, m$

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} $
$B=5 \times 10^{-4} T$