Question

A wire carries a steady current of $50\text{A}$ as shown in the figure. Find the magnetic field caused by a $1\text{cm}$ segment of this wire at a point $\text{P}$, which is $1\text{m}$ away.

• A. $30\text{nT},\text{inward}$
• B. $50\text{nT},\text{outward}$
• C. $30\text{nT},\text{outward}$
• D. $30\text{nT},\text{inward}$

Answer 1

The correct option is D $30\text{nT},\text{inward}$


From Biot- Savart's law,

$\overrightarrow{B}=\frac{{\mu }_{0}I}{4\pi }\frac{(\overrightarrow{dl}\times \overrightarrow{r})}{r^3}$

$B=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$

Since, current carrying wire segment $\left(dl\right)$ is very small as compare to distance between wire and point $\text{P}$, so we can apply above formula directly for given segment.

Given:
$I=50\text{A};r=1\text{m};dl=1\text{cm}={10}^{-2}\text{m}$

$B=\frac{4\pi \times {10}^{-7}}{4\pi }\times \frac{50\times {10}^{-2}\sin {90}^{\circ }}{1^2}$

$B=50\times {10}^{-9}\text{T}$

From the right-hand thumb rule, it is clear that the direction of the magnetic field at point $\text{P}$ is perpendicularly inward on the plane of paper.

$\therefore B=50\text{nT , inward}(\otimes )$

Hence, option $\left(D\right)$ is the correct answer.