Induced Current Question

1. Apr 22, 2017

BrainMan

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution

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

2. Apr 22, 2017

TSny

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.

3. Apr 22, 2017

BrainMan

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

So like this?

4. Apr 22, 2017

TSny

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

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

5. Apr 23, 2017

BrainMan

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

6. Apr 23, 2017

TSny

Looks good.

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