Calculating change in current in a solenoid

[redsox133]

Homework Statement

A long, thin solenoid has 400 turns per meter and a radius of 2.65 cm. The current in the solenoid is increasing at a uniform rate dI/dt, and the induced electric field at a point near the center of the solenoid and 3.25 cm from its axis is 5.00✕ 10-6 V/m. Calculate dI/dt.

Homework Equations

$B_s = u_onI$
$N\frac{d\phi}{dt}=M\frac{dI}{dt}=\mathcal{EMF}$
$\oint \vec E \cdot d \vec L = \frac{dB}{dt}$

The Attempt at a Solution

I tried solving for $\frac{dB}{dt}$ using the third equation. I ended up with $E=\frac{r}{2}\frac{dB}{dt}$, and then I plugged in E and r to solve for $\frac{dB}{dt}$. I wanted to use that to solve for $\frac{dI}{dt}$ in the second equation but I don't know what M is. After that I tried taking the derivative of the first equation to get $\frac{dB}{dt}=u_on\frac{dI}{dt}$ and then solving for $\frac{dI}{dt}$ but I got the wrong answer. If someone could let me know if I'm on the right track I would appreciate it.

 

[anathema]

Hello there,

Your approach is on the right track, but there are a few errors in your calculations. Let's go through them step by step.

First, in the third equation, the integral should be taken over a closed path, not just a straight line. This means that you need to take into account the radius of the solenoid and the distance from the axis in your calculation. The correct equation should be:

$\oint \vec E \cdot d \vec L = \frac{\pi r^2}{2}\frac{dB}{dt}$

Next, when you plug in the given values for the electric field and the radius, you should get:

$5.00 \times 10^{-6} \text{V/m} = \frac{\pi (0.0325 \text{m})^2}{2}\frac{dB}{dt}$

Solving for $\frac{dB}{dt}$, we get:

$\frac{dB}{dt} = \frac{5.00 \times 10^{-6} \text{V/m}}{\pi (0.0325 \text{m})^2 / 2} = 3.06 \times 10^{-3} \text{T/s}$

This is the correct value for the rate of change of the magnetic field inside the solenoid.

Now, let's move on to the second equation. The value of M is the mutual inductance between the solenoid and the induced electric field. In this case, it is equal to the self-inductance of the solenoid, which can be calculated using the first equation:

$B_s = \mu_0 n I$

Rearranging this equation, we get:

$I = \frac{B_s}{\mu_0 n}$

Substituting this value of I into the second equation, we get:

##N\frac{d\phi}{dt} = M\frac{dB}{dt} = \frac{\mu_0 n \pi r^2}{2}\frac{dB}{dt} = \frac{\mu_0 n \pi r^2}{2}\frac{5.00 \times 10^{-6} \text{T/s}}{\pi (0.0325 \text{m})^2 / 2} = 3.06 \times 10^{-3} \text