Induced Current in a Coil with Changing Magnetic Field

[BrainMan]

Homework Statement

A 5.0-cm-diameter coil has 20 turns and a resistance of 0.50Ω. A magnetic field perpendicular to the coil is B=0.020t+0.010t^2, where B is in tesla and t is in seconds.

Find an expression for the induced current I(t) as a function of time.

Homework Equations

The Attempt at a Solution

$$\varepsilon = \frac{d\phi}{dt}$$
$$\varepsilon dt = d\phi$$
$$\int\varepsilon dt = \int d\phi$$
$$\phi = (0.020t + .010t^2) * \pi r^2$$
$$\varepsilon \int dt = \int d\phi$$
$$\varepsilon t = \pi r^2 [ (1/2) (0.020t^2) + (1/3)(.010t^3)]$$
$$\varepsilon = \frac {I}{R}$$
$$I = \frac{\pi r^2 [ (1/2) (0.020t^2) + (1/3)(.010t^3)]}{tR}$$

 

[TSny]

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

OK, except you haven't taken into account that you have 20 turns in the coil.

$$\varepsilon dt = d\phi$$
$$\int\varepsilon dt = \int d\phi$$

Don't do this.

Use your expression $\phi = (0.020t + .010t^2) * \pi r^2$ in $\varepsilon = \frac{d\phi}{dt}$ along with the correction for the 20 turns.

 

[BrainMan]

OK, except you haven't taken into account that you have 20 turns in the coil.Don't do this.

Use your expression $\phi = (0.020t + .010t^2) * \pi r^2$ in $\varepsilon = \frac{d\phi}{dt}$ along with the correction for the 20 turns.

So $$I = \frac{20\pi r^2 [ (1/2) (0.020t^2) + (1/3)(.010t^3)]}{R}$$

So like this?

 

[TSny]

So $$I = \frac{20\pi r^2 [ (1/2) (0.020t^2) + (1/3)(.010t^3)]}{R}$$

So like this?

No. The numerator in your expression for $I$ does not equal $\varepsilon$.

If $\phi = (0.020t + .010t^2) * \pi r^2$, then what is $\frac{d\phi}{dt}$?

 

[BrainMan]

No. The numerator in your expression for $I$ does not equal $\varepsilon$.

If $\phi = (0.020t + .010t^2) * \pi r^2$, then what is $\frac{d\phi}{dt}$?

Oh ok I get it now. The answer should be $$\frac {20\pi r^2(0.020+.020t)}{R}$$

 

[TSny]

The answer should be $$\frac {20\pi r^2(0.020+.020t)}{R}$$

Looks good.