Solving an Ideal Solenoid: Calculating Magnetic Field, Self-Inductance, and RMS Current

[Saketh]

Homework Statement

A wire of length $\ell$, radius $r_1$, and resistivity $\rho$ is tightly wound in a single layer into the shape of a solenoid with circular cross section of radius $r_2$. Assume it is an ideal solenoid. A DC voltage $V$ is placed across the ends of the solenoid.

a. What is the magnitude of the magnetic field inside the solenoid?
b. What is the self-inductance of the solenoid?
c. If the DC voltage source is replaced with an AC source with rms voltage $V_{rms}$ and frequency $f$, what is the rms current $I_{rms}$ through the solenoid?

Homework Equations

$$R = \frac{\rho \ell}{A}$$
Ampere's law
Ohm's law

The Attempt at a Solution

a.

I found current and turn density in terms of the given variables.

$$I = \frac{V}{R} = \frac{\pi V r_1^2}{\rho \ell}$$
$$N = \frac{\ell}{2 \pi r_2}$$

I plugged everything into the equation for resistance, and got the following expression for magnetic field.

$$B = \frac{\mu_0 V r_1^2}{2 \rho \ell r_2}$$

b.

$$\mathrm{Magnetic flux linkage} = N\Phi = LI$$
$$L = \frac{ABN}{I}$$
Plugging in, I got
$$L = \frac{\mu_0 \ell}{2 \pi}$$.

c.

$$V(t) = L\frac{dI}{dt}$$

$$V_{rms}\sqrt{2} \sin{2\pi f t} dt = L dI$$

$$-\frac{V_{rms}\cos{2\pi f t}}{\sqrt{2} \pi f} = L I(t)$$

$$I_{rms} = \frac{2V_{rms}}{\mu_0 f \ell}$$

I have absolutely no idea if I did any of this problem correctly, so I would appreciate it if someone could check what I did.

 

[anathema]

Hello,

I would like to offer some feedback on your solution attempt.

For part a, you correctly found the expression for current and turn density. However, when plugging them into the equation for resistance, there is an error. The correct expression for the magnetic field inside the solenoid is:

B = \frac{\mu_0 NI}{L}

Where N is the number of turns and I is the current. Also, since the solenoid is an ideal solenoid, the resistivity term should not be included in the equation.

For part b, you correctly found the expression for self-inductance, but there is a minor error in your final answer. The correct expression for self-inductance is:

L = \frac{\mu_0 N^2 A}{\ell}

Where A is the cross-sectional area of the solenoid.

For part c, you correctly set up the equation for V(t), but there is an error in the final expression. The correct expression for I_{rms} is:

I_{rms} = \frac{V_{rms}}{2\pi f L}

Overall, your solution attempt was on the right track, but there were some minor errors that can be easily corrected. Keep up the good work!

 

[m2003]

Overall, your attempt at solving the problem seems to be on the right track. However, there are a few things that could be clarified or corrected.

a. Your expression for current, I, is correct. However, the expression for turn density, N, should be N = \frac{\pi r_2^2}{\ell} instead of N = \frac{\ell}{2 \pi r_2}. This is because turn density is defined as the number of turns per unit length, so you need to divide by the length of one turn, which is the circumference of the solenoid, 2 \pi r_2.

Also, your expression for magnetic field, B, is missing a factor of N. It should be B = \frac{\mu_0 NI}{\ell}.

b. Your expression for self-inductance, L, is incorrect. The correct expression is L = \frac{\mu_0 N^2 \pi r_2^2}{\ell}. This can be obtained by substituting the expressions for turn density and current into the formula L = \frac{ABN}{I}.

c. Your expression for the rms current, I_{rms}, is correct. However, it is missing a factor of \sqrt{2} in the denominator. It should be I_{rms} = \frac{2V_{rms}}{\mu_0 f \ell \sqrt{2}}.

Overall, your approach to the problem is correct. Just make sure to double check your calculations and pay attention to the units. Also, it may be helpful to label your final answers with appropriate units (e.g. Tesla for magnetic field, Henry for self-inductance, and Ampere for current).