Question

Figure (36-E1) shows some of the equipotential surfaces of the magnetic scalar potential. Find the magnetic field B at a point in the region.
Figure

Answer 1

Explanation:

Given data in the question  

Perpendicular distance, $d x=10 \sin 30^{\circ} \mathrm{cm}=0.05 \mathrm{m}$

potential changes, $d V=0.1 \times 10^{-4} T-m$

We know the relationship between the potential and the field is established by

$B=-\frac{d V}{d x}$

$B=\frac{\left(-0.1 \times 10^{-4} T-m\right)}{(5 \times 10-2 m)}$

$B=\frac{-1 \times 10^{-5}}{5 \times 10^{-2}}$

$B=\frac{-1 \times 10^{-3}}{5}$

$\begin{aligned}&B=-0.2 \times 10^{-3}\\&B=-2 \times 10^{-4} T\end{aligned}$

B is perpendicular to surface equipotential. The positive x-axis is at an angle of 120 ° here.