Question

Figure shows some of the equipotential surfaces of the magnetic scalar potential. Find the magnetic field B at a point in the region.

Answer 1

Here x = 10 sin 30 cm

= 5 cm = 0.05 m,

and dv = $0.1\times {10}^{-4}$ T-m

Since B = $\frac{dv}{dx}$

= $\frac{0.1\times {10}^{-4}\times T-m}{5\times {10}^{-2}m}$

= $2\times {10}^{-4}$ T.

Since B is perpendicular to equipotential surface. Here it is at angle ${120}^0$ with (+ve) x-axis.