JEE MAIN - Physics (2022 - 26th June Morning Shift - No. 11)
The magnetic flux through a coil perpendicular to its plane is varying according to the relation $$\phi = (5{t^3} + 4{t^2} + 2t - 5)$$ Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2 s will be,
15.6 A
16.6 A
17.6 A
18.6 A
Explanation
$\phi=5 \mathrm{t}^3+4 \mathrm{t}^2+2 \mathrm{t}-5$
$|\mathrm{e}|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=15 \mathrm{t}^2+8 \mathrm{t}+2$
At $\mathrm{t}=2,|\mathrm{e}|=15 \times 2^2+8 \times 2+2$
$\Rightarrow \mathrm{e}=78 \mathrm{~V} $
$\Rightarrow \mathrm{I}=\frac{\mathrm{e}}{\mathrm{R}}=\frac{78}{5}=15.60$
$|\mathrm{e}|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=15 \mathrm{t}^2+8 \mathrm{t}+2$
At $\mathrm{t}=2,|\mathrm{e}|=15 \times 2^2+8 \times 2+2$
$\Rightarrow \mathrm{e}=78 \mathrm{~V} $
$\Rightarrow \mathrm{I}=\frac{\mathrm{e}}{\mathrm{R}}=\frac{78}{5}=15.60$
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