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Question
Figure shows some of the equipotential surfaces of the magnetic scalar potential. Find the magnetic field B at a point in the region.
Solution
Given :
Perpendicular distance, `dx = 10 sin 30^circ "cm" = 0.05 "m"` ,
Change in the potential, dV = `0.1 xx 10^-4 "T-m"`
We know that the relation between the potential and the field is given by
`B = -(dV)/(dx)`
⇒ `B = -(0.1 xx 10^-4 "T-m")/(5 xx 10^-2 "m")`
⇒ `B = -2 xx 10^-4 "T"`
B is perpendicular to the equipotential surface. Here, it is at angle of 120° with the positive x-axis.
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