Q. 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 $\left( \mu_{0} = 4\pi \times 10^{- 7}\frac{H}{m} \right)$
Explanation
(c) $\ i = 30\ A,\ B = 4 \times 10^{- 4}\ T$
$r = 2\ cm = 2 \times 10^{- 2}\ m$
Initial magnetic field (parallel to the wire)
$B_{1} = 4 \times 10^{- 4}\ T$
Magnetic field produced by the straight wire
$B_{2} = \frac{\mu_{0}i}{2\pi r} = \frac{2 \times 10^{- 7} \times 30}{2 \times 10^{- 2}}$
$\ \ \ \ \ \ = 3 \times 10^{- 4}\ T$
$B_{2}$ will be in the plane perpendicular to the plane of wire, so
$B_{1}$ and $B_{2}$ are perpendicular to each other.
$\therefore$ Resultant magnetic field
$B = \ \sqrt{B_{1}^{2} + B_{2}^{2}}$
$\ \ \ \ \ = \sqrt{\left( 4 \times 10^{- 4} \right)^{2} + \left( 3 \times 10^{- 4} \right)^{2}}$
$\ \ \ \ \ = 5 \times 10^{- 4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.