JEE Main 2014 · 19 Apr · Shift 1 · Physics Q21

← JEE Main 2014 · 19 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 21 of 90

Figure shows a circular area of radius R where a uniform magnetic field $\vec B$ is going into the plane of paper and increasing in magnitude at a constant rate. In that case, which of the following graphs, drawn schematically, correctly shows the variation of the induced electric field E(r) ?

A circle of radius R filled with crosses (magnetic field into the page).

E rises linearly from 0 to a peak at R, then falls off as a curve

Graph of E against r with R marked on the r-axis: a straight line from the origin up to a peak at R, then a decreasing curve.

E rises linearly to a peak at R, then falls linearly

Graph of E against r with R marked on the r-axis: a straight rise to a peak at R, then a straight-line fall.

E rises linearly to R, then stays constant

Graph of E against r with R marked on the r-axis: a straight rise from the origin up to R, then a horizontal line.

E rises along a curve to a sharp peak at R, then falls to a minimum and rises slightly

Graph of E against r with R marked on the r-axis: a concave-up rise to a cusp at R, then a fall to a shallow minimum.

Answer and worked solution

Official answer: A

By Faraday's law round a circle of radius $r$ : $E\cdot2\pi r=\pi r^2\dfrac{dB}{dt}$ inside, so $E\propto r$ for $r<R$ .

Outside, $E\cdot2\pi r=\pi R^2\dfrac{dB}{dt}$ , so $E\propto\dfrac1r$ . That is a linear rise to a peak at R, then a $\frac1r$ fall: graph (1).

Topics:Faraday's and Lenz's laws

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JEE Main 2014 · 19 Apr · Shift 1, Physics question 21