JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 23)
A very long solenoid of radius R is carrying
current I(t) = kte–at(k > 0), as a function of time
(t $$ \ge $$ 0). counter clockwise current is taken to be
positive. A circular conducting coil of radius
2R is placed in the equatorial plane of the
solenoid and concentric with the solenoid. The
current induced in the outer coil is correctly
depicted, as a function of time, by :-
_9th_April_Evening_Slot_en_23_1.png)
_9th_April_Evening_Slot_en_23_2.png)
_9th_April_Evening_Slot_en_23_3.png)
_9th_April_Evening_Slot_en_23_4.png)
Explanation
$$\varepsilon = - {{d\phi } \over {dt}} = {{ - d} \over {dt}}\left( {{\mu _0}nl} \right)\left( {\pi {R^2}} \right)$$
$$ = - \left( {{\mu _0}n\pi {R^2}} \right){{dl} \over {dt}}$$
$$ = - {\mu _0}n\pi {R^2}{{d\left( {kt{e^{ - \omega t}}} \right)} \over {dt}}$$
$$ = {\rm{ }}-C{\rm{ }}\left( {t\left( {{e^{-\omega t}}{\rm{ }}} \right){\rm{ }}\left( {-\alpha } \right){\rm{ }} + {\rm{ }}{e^{-\omega t}}{\rm{ }}} \right){\rm{ }}\left( {C{\rm{ }} \to {\rm{ }}constant} \right)$$
= –C (e–$$\omega $$t) (1 – $$\alpha $$t)
Clearly, at t = 0
Induced current $$ \ne $$ 0
Also, apply Lenz law to find correct option
$$ = - \left( {{\mu _0}n\pi {R^2}} \right){{dl} \over {dt}}$$
$$ = - {\mu _0}n\pi {R^2}{{d\left( {kt{e^{ - \omega t}}} \right)} \over {dt}}$$
$$ = {\rm{ }}-C{\rm{ }}\left( {t\left( {{e^{-\omega t}}{\rm{ }}} \right){\rm{ }}\left( {-\alpha } \right){\rm{ }} + {\rm{ }}{e^{-\omega t}}{\rm{ }}} \right){\rm{ }}\left( {C{\rm{ }} \to {\rm{ }}constant} \right)$$
= –C (e–$$\omega $$t) (1 – $$\alpha $$t)
Clearly, at t = 0
Induced current $$ \ne $$ 0
Also, apply Lenz law to find correct option
Comments (0)
