Question

A long straight vertical conductor carries a current of $8A$ in the upward direction. What is the magnitude of the resultant magnetic induction at a point in the horizontal plane at a distance of $4cm$ from the conductor towards South?
(The horizontal component of earth's magnetic induction = $4\times {10}^{-5}T$)

• A. $2\sqrt{2}\times {10}^{-5}T$
• B. $4\sqrt{2}\times {10}^{-5}T$
• C. $4\times {10}^{-5}T$
• D. $2\times {10}^{-5}T$

Answer 1

Answer: (B)

$4\sqrt{2}\times {10}^{-5}T$

This field due to the coil will be in a direction west to east which is perpendicular to the direction of earth's magnetic field. Let east be positive x-direction and north be positive y-axis.

Magnetic field due to a long current carrying conductor is given by the formula,

$B_l=\frac{{\mu }_{0}I}{2\pi a}$

${\overrightarrow{B}}_l=\frac{8\times 4\pi \times {10}^{-7}}{2\pi \times \left(0.04\right)}=4\times {10}^{-5}T\hat{i}$

Magnetic field due to earth's horizontal magnetic field is:

${\overrightarrow{B}}_h=4\times {10}^{-5}T\hat{j}$

Hence, resultant magnetic field is:

$\overrightarrow{B}={\overrightarrow{B}}_l+{\overrightarrow{B}}_h$

$|\overrightarrow{B}|=\sqrt{(4\times {10}^{-5}{)}^2+(4\times {10}^{-5}{)}^2}=4\sqrt{2}\times {10}^{-5}T$