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In a coil of resistance 10 $\Omega$, the induced current developed by changing magnetic flux through it, is shown in figure as a function of time. The magnitude of change in flux through the coil in Weber is
A
4
B
8
C
2
D
6
Detailed Solution
$\left|\frac{d\phi}{dt}\right| = e = iR \Rightarrow d\phi = (iR)\,dt$
$\Delta\phi = R\int i\,dt = R\times$ (area under the i–t graph)
The graph is a triangle with height 4 A and base 0.1 s: area $= \frac{1}{2}\times4\times0.1 = 0.2$ A s
$\Delta\phi = 10\times0.2 = 2$ Wb
$\Delta\phi = R\int i\,dt = R\times$ (area under the i–t graph)
The graph is a triangle with height 4 A and base 0.1 s: area $= \frac{1}{2}\times4\times0.1 = 0.2$ A s
$\Delta\phi = 10\times0.2 = 2$ Wb