JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 15)
A long solenoid of radius R carries a time (t) - dependent current
I(t)=I0t(1 - t). A ring of radius 2R is placed coaxially near its middle. During the time interval 0 $$ \le $$ t $$ \le $$ 1, the induced current (IR) and the induced EMF(VR) in the ring change as :
I(t)=I0t(1 - t). A ring of radius 2R is placed coaxially near its middle. During the time interval 0 $$ \le $$ t $$ \le $$ 1, the induced current (IR) and the induced EMF(VR) in the ring change as :
Direction of IR remains unchanged and VR is zero at t = 0.25
Direction of IR remains unchanged and VR is maximum at t = 0.5
At t = 0.25 direction of IR reverses and VR is maximum
At t = 0.5 direction of IR reverses and VR is zero
Explanation
I(t) = I0t(1 - t)
We know, $$\phi $$ = BA
$$ \Rightarrow $$ $$\phi $$ = $$\mu $$0nIA
$$ \Rightarrow $$ $$\phi $$ = $$\mu $$0nAI0(t - t2)
Also VR = $$ - {{d\phi } \over {dt}}$$
= - $$\mu $$0nAI0(1 - 2t)
VR = 0 when 1 - 2t = 0
$$ \Rightarrow $$ t = 0.5
Also we know, VR = IRr
$$ \Rightarrow $$ IR = $${{{\mu _0}nA{I_0}\left( {1 - 2t} \right)} \over r}$$
after t = 0.5, IR reverses its direction.
We know, $$\phi $$ = BA
$$ \Rightarrow $$ $$\phi $$ = $$\mu $$0nIA
$$ \Rightarrow $$ $$\phi $$ = $$\mu $$0nAI0(t - t2)
Also VR = $$ - {{d\phi } \over {dt}}$$
= - $$\mu $$0nAI0(1 - 2t)
VR = 0 when 1 - 2t = 0
$$ \Rightarrow $$ t = 0.5
Also we know, VR = IRr
$$ \Rightarrow $$ IR = $${{{\mu _0}nA{I_0}\left( {1 - 2t} \right)} \over r}$$
after t = 0.5, IR reverses its direction.
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