Question

In the given figure the magnetic flux through the loop increases according to the relation $\phi_B(t) = 10t^2 + 20t$, where $\phi_B$ is in milliwebers and t is in seconds. The magnitude of current through R=2.0 $\Omega$ resistor at t=5 s is ________ mA. 

 

Hint:

Be extremely careful with units. Since the magnetic flux was given in milliwebers (mWb), the calculated EMF from its time derivative will be in millivolts (mV). This directly gives the current in milliamperes (mA) when divided by resistance in ohms.

Answer 1

Correct Answer: 60

According to Faraday's law of electromagnetic induction, the magnitude of the induced electromotive force (EMF), $|\mathcal{E}|$, is equal to the rate of change of magnetic flux.
$|\mathcal{E}| = \left|\frac{d\phi_B}{dt}\right|$.
The magnetic flux is given as $\phi_B(t) = 10t^2 + 20t$ in units of milliwebers (mWb).
First, find the derivative of the flux with respect to time.
$\frac{d\phi_B}{dt} = \frac{d}{dt}(10t^2 + 20t) = 20t + 20$.
Since $\phi_B$ is in mWb, the induced EMF $\mathcal{E}$ will be in millivolts (mV).
$|\mathcal{E}| = (20t + 20)$ mV.
We need to find the EMF at $t = 5$ s.
$|\mathcal{E}|_{t=5s} = 20(5) + 20 = 100 + 20 = 120$ mV.
Now, using Ohm's law, the current $I$ through the resistor R is given by $I = \frac{|\mathcal{E}|}{R}$.
We have $|\mathcal{E}| = 120$ mV and $R = 2.0 \ \Omega$.
$I = \frac{120 \text{ mV}}{2.0 \ \Omega} = 60$ mA.
The magnitude of the current is 60 mA.