Question

An element $△l=△x\overline{l}$ is placed at the origin and carries and carries a current l=10A along x-axis.
If $△x=1cm$, magnetic field at point P is :

• A. $4\times {10}^{-8}\hat{k}T$
• B. $4\times {10}^{-8}\hat{i}T$
• C. $4\times {10}^{-8}\hat{j}T$
• D. $-4\times {10}^{-8}\hat{j}T$

Answer 1

The correct option is C $4\times {10}^{-8}\hat{j}T$

$\text{Given}:$ $\Delta \ell =\Delta x={10}^{–2}m$

I = 10A

r = 0.5 m

$\text{Solution}:$

We know,

$\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }\left[\frac{Id\ell \times r}{r^3}\right]$

=$\left[\frac{{\mu }_{0}I\Delta \ell sin\theta }{4\pi \times r^2}\right]$

$\therefore$ magnetic field on y axis = $\left[\frac{4\pi \times {10}^{–7}}{4\pi }\right]\times \left[\frac{10\times {10}^{–2}}{(0.5{)}^2}\right]$

= $\left[\frac{{10}^{–8}}{25\times {10}^{–2}}\right]$

= $0.04\times {10}^{–6}=4\times {10}^{–8}T$

i.e. $dB=4\times {10}^{-8}\hat{j}T$ (y axis)

$\text{Hence C is the correct option}$