JEE MAIN - Physics (2021 - 25th July Morning Shift - No. 20)
A circular conducting coil of radius 1 m is being heated by the change of magnetic field $$\overrightarrow B $$ passing perpendicular to the plane in which the coil is laid. The resistance of the coil is 2 $$\mu$$$$\Omega$$. The magnetic field is slowly switched off such that its magnitude changes in time as
$$B = {4 \over \pi } \times {10^{ - 3}}T\left( {1 - {t \over {100}}} \right)$$
The energy dissipated by the coil before the magnetic field is switched off completely is E = ___________ mJ.
$$B = {4 \over \pi } \times {10^{ - 3}}T\left( {1 - {t \over {100}}} \right)$$
The energy dissipated by the coil before the magnetic field is switched off completely is E = ___________ mJ.
Answer
80
Explanation
$$\phi = \overrightarrow B .\overrightarrow S $$
$$\phi = {4 \over \pi } \times {10^{ - 3}}\left( {1 - {t \over {100}}} \right).\pi {R^2}$$
$$\phi = 4 \times {10^{ - 3}} \times {(1)^2}\left( {1 - {t \over {100}}} \right)$$
$$\varepsilon = {{ - d\phi } \over {dt}}$$
$$\varepsilon = {{ - d} \over {dt}}\left( {4 \times {{10}^{ - 3}}\left( {1 - {t \over {100}}} \right)} \right)$$
$$\varepsilon = 4 \times {10^{ - 3}}\left( {{1 \over {100}}} \right) = 4 \times {10^{ - 5}}V$$
When B = 0
$$1 - {t \over {100}} = 0$$
t = 100 sec
Heat $$ = {{{\varepsilon ^2}} \over R}t$$
Heat $$ = {{{{(4 \times {{10}^{ - 5}})}^2}} \over {2 \times {{10}^{ - 6}}}} \times 100$$ J
Heat $$ = {{16 \times {{10}^{ - 10}} \times 100} \over {2 \times {{10}^{ - 6}}}}$$ J
Heat = 0.08 J
Heat = 80 mJ
$$\phi = {4 \over \pi } \times {10^{ - 3}}\left( {1 - {t \over {100}}} \right).\pi {R^2}$$
$$\phi = 4 \times {10^{ - 3}} \times {(1)^2}\left( {1 - {t \over {100}}} \right)$$
$$\varepsilon = {{ - d\phi } \over {dt}}$$
$$\varepsilon = {{ - d} \over {dt}}\left( {4 \times {{10}^{ - 3}}\left( {1 - {t \over {100}}} \right)} \right)$$
$$\varepsilon = 4 \times {10^{ - 3}}\left( {{1 \over {100}}} \right) = 4 \times {10^{ - 5}}V$$
When B = 0
$$1 - {t \over {100}} = 0$$
t = 100 sec
Heat $$ = {{{\varepsilon ^2}} \over R}t$$
Heat $$ = {{{{(4 \times {{10}^{ - 5}})}^2}} \over {2 \times {{10}^{ - 6}}}} \times 100$$ J
Heat $$ = {{16 \times {{10}^{ - 10}} \times 100} \over {2 \times {{10}^{ - 6}}}}$$ J
Heat = 0.08 J
Heat = 80 mJ
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