In the given figure the magnetix flux through the loop increases according to the relation - Bt-10t2-20t- where -B is in milliwebers and t is in seconds-The magnitude of current through R-2 - resistor at t-5 s is in mA

As we know that, magnitude of EMF is given by | ϵ nbsp; |=dϕdt=d(10t^2+20t)dt=(20t+20) mV And, magnitude of current is given as | i |= | ϵ nbsp; |R=20t+202 ...

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Byju's Answer

Standard XII

Physics

Dependence on Magnetic Field

In the given ...

Question

In the given figure the magnetix flux through the loop increases according to the relation $ϕ_{B} (t) = 10 t^{2} + 20 t$ , where $ϕ_{B}$ is in milliwebers and $t$ is in seconds.
The magnitude of current through $R = 2 Ω$ resistor at $t = 5 s$ is $(in mA)$

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Solution

As we know that, magnitude of EMF is given by
$| ϵ | = \frac{d ϕ}{d t} = \frac{d (10 t^{2} + 20 t)}{d t} = (20 t + 20) mV$
And, magnitude of current is given as
$| i | = \frac{| ϵ |}{R} = \frac{20 t + 20}{2} = (10 t + 10) mA$
Thus, at $t = 5 seconds$
$| i | = 10 \times 5 + 10 = 60 mA$

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In the given figure the magnetix flux through the loop increases according to the relation ϕB(t)=10t2+20t, where ϕB is in milliwebers and t is in seconds. The magnitude of current through R=2 Ω resistor at t=5 s is (in mA)

As we know that, magnitude of EMF is given by |ϵ|=dϕdt=d(10t2+20t)dt=(20t+20) mV And, magnitude of current is given as |i|=|ϵ|R=20t+202=(10t+10) mA Thus, at t=5 seconds |i|=10×5+10=60 mA

In the given figure the magnetix flux through the loop increases according to the relation $ϕ_{B} (t) = 10 t^{2} + 20 t$ , where $ϕ_{B}$ is in milliwebers and $t$ is in seconds.
The magnitude of current through $R = 2 Ω$ resistor at $t = 5 s$ is $(in mA)$

As we know that, magnitude of EMF is given by
$| ϵ | = \frac{d ϕ}{d t} = \frac{d (10 t^{2} + 20 t)}{d t} = (20 t + 20) mV$
And, magnitude of current is given as
$| i | = \frac{| ϵ |}{R} = \frac{20 t + 20}{2} = (10 t + 10) mA$
Thus, at $t = 5 seconds$
$| i | = 10 \times 5 + 10 = 60 mA$