JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 25)
In a coil of resistance $$8 \,\Omega$$, the magnetic flux due to an external magnetic field varies with time as $$\phi=\frac{2}{3}\left(9-t^{2}\right)$$. The value of total heat produced in the coil, till the flux becomes zero, will be _____________ $$J$$.
Answer
2
Explanation
$$R = 8\,\Omega $$
$$\phi = {2 \over 3}(9 - {t^2})$$
At $$t = 3$$, $$\phi = 0$$
$$\varepsilon = \left| { - {{d\phi } \over {dt}}} \right| = {4 \over 3}t$$
$$H = \int_0^3 {{{{V^2}} \over R}dt = \int_0^3 {{1 \over 8} \times {{16} \over 9}{t^2}dt} } $$
$$ = {2 \over 9} \times \left( {{{{t^3}} \over 3}} \right)_0^3 = {2 \over {9 \times 3}} \times 27 = 2\,J$$
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