JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 11)
The time taken for the magnetic energy to reach 25% of its maximum value, when a solenoid of resistance R, inductance L is connected to a battery, is :
infinite
$${L \over R}$$ ln10
$${L \over R}$$ ln2
$${L \over R}$$ ln5
Explanation
Magnetic energy, U = $${1 \over 2}$$LI$$_0^2$$
Given : U = 25% of U0.
$$ \Rightarrow $$ $${1 \over 2}L{I^2} = {1 \over 4} \times {1 \over 2}LI_0^2$$
$$ \Rightarrow {I^2} = {{I_0^2} \over 4} \Rightarrow I = {{{I_0}} \over 2}$$
$$ \therefore $$ $$I = {I_0}(1 - {e^{ - t/\tau }})$$
$$ \Rightarrow {{{I_0}} \over 2} = {I_0}(1 - {e^{ - t/\tau }})$$
$$ \Rightarrow {1 \over 2} = {e^{ - t/\tau }}$$
$$ \Rightarrow {e^{t/\tau }} = 2$$
$$ \Rightarrow t = \tau \ln 2$$
$$ \Rightarrow t = {L \over R}\ln 2$$
Given : U = 25% of U0.
$$ \Rightarrow $$ $${1 \over 2}L{I^2} = {1 \over 4} \times {1 \over 2}LI_0^2$$
$$ \Rightarrow {I^2} = {{I_0^2} \over 4} \Rightarrow I = {{{I_0}} \over 2}$$
$$ \therefore $$ $$I = {I_0}(1 - {e^{ - t/\tau }})$$
$$ \Rightarrow {{{I_0}} \over 2} = {I_0}(1 - {e^{ - t/\tau }})$$
$$ \Rightarrow {1 \over 2} = {e^{ - t/\tau }}$$
$$ \Rightarrow {e^{t/\tau }} = 2$$
$$ \Rightarrow t = \tau \ln 2$$
$$ \Rightarrow t = {L \over R}\ln 2$$
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