Question

The magnetic flux through a metal ring varies with time t according to ΦB = at3 − bt2, where ΦB is in webers, a = 6.00 Wb s−3, b = 18.0 Wb s−2, and t is in seconds. The resistance of the ring is 2.80 Ω. For the interval from t = 0 to t = 2.00 s, determine the maximum current induced in the ring.

Answer 1

To solve this problem it is necessary to apply the second derivative of the function to find the maximum time reference and thus calculate the maximum voltage.

With the maximum voltage by Ohm's Law it is possible to find the maximum current.

Ohm's law defines that

E = I*R

Where,

I = Current

R= Resistance

On the other hand by faraday studies and the potential can be expressed at the rate of change of the electric flow, that is

$E = \frac{d\phi}{dt}$

Replacing with our values we have that

$E = \frac{d(6t^3-18t^2)}{dt}$

$E = -18t^2 +36t$

The second derivative is

$E' = -36t+36$

When E' = 0 we have a Maximum, then

0 = -36t+36

t = 1

Therefore when the time is 1s E has a Maximum, replacing at the function

$E(t) = -18t^2 +36t$

$E(1) = -18(1)^2 +36(1)$

$E = 18V$

Then the maximum current will be given by

$I = \frac{E}{R}$

$I = \frac{18}{2.8}$

$I = 6.42A$

Therefore the maximum current induced in the ring is 6.42A