An ideal solenoid with diameter d 138 times 10 1 m and...

An ideal solenoid with diameter $D = 1.38 \times 10^{-1}$ m and length $L = 2.70 \times 10^{-1}$ m is connected to a voltage source. A smaller solenoid, with diameter $\frac{D}{3}$ and length $\frac{L}{3}$, is placed inside the first. Both solenoids have the same number of turns per unit length $n = 5500$. a) Find the magnitude of the magnetic field due to the large Solenoid when $I = 4.00$ A. $|\vec{B}| = 2.76 \times 10^{-2}$ c) What is the magnetic flux through the small solenoid due to the large solenoid? $\Phi = 4.58 \times 10^{-5}$ d) What would be the induced emf in the coil if the current in the solenoid is reduced to zero in $t = 5.70 \times 10^{-2}$ s? $\xi = 0.397$ The current in the solenoid now changes as $i(t) = (4.00 \text{ A}) \sin(160 \cdot t)$. f) Find the maximum magnitude of the magnetic field produced by the large solenoid: $B_{\text{max}} = 2.76 \times 10^{-2}$ g) Find the maximum magnitude of the magnetic flux through the small solenoid. $\Phi_{\text{max}} = 4.58 \times 10^{-5}$ h) Find the maximum magnitude of the induced Emf (voltage) around the small solenoid. $\xi_{\text{max}} = 4.57$ i) The induced emf in the small solenoid at $t = 1.38 \times 10^{-1}$ s. $\xi(t = 1.38 \times 10^{-1} \text{ s}) = 3.40$ ```

Transcript

00:01 Hello, in this question we're told we have a solenoid, and this solenoid has 200 wraps.

00:05 We're given the radius, the length, and the current running in the solenoid, and we're asked various things.

00:10 In part a, we're asked what is the magnitude of the uniform magnetic field within the solenoid? so what is b? well, we know that b inside of a solenoid is going to be mu -naught times our number of wraps per unit length times the current in the wire of that solenoid.

00:27 And we have all those values, so we can just go ahead and plug them in.

00:30 If i do that, i will get a magnetic field of 0 .0623 teslas, which i can also write as 62 .3 millitesla, which is one of our options, so that's a good sign.

00:46 Part b then says, consider a loop, a radius 2r that shares a center point with the solenoid.

00:53 So i'm going to have another loop out here with a radius of 2r for that loop.

01:02 And we're asked, what is the magnetic flux through the plane of this circular conducting loop in weber's? so what is our magnetic flux in that loop? well, we know that magnetic flux is definitionally b times a, and technically it's b a cosine theta, but in this case, our area vector and our magnetic field vector are both pointing out of the page, so we don't have to worry about any angle stuff.

01:25 So we know the b, well, the b inside is going to be 62 .3 millitesla.

01:32 And then our area is just going to be pi times our radius, which is 2r squared there.

01:40 And that will give us a value of flux of 2 .81 times 10 to the negative 3, so microweber's, or that's 2 .82, i had a little rounding error there.

01:56 And so that will be our last option in our list of options.

01:58 And then we're asked in part c, if we were to change the current in the solenoid, so we have an initial current of, we'll say, 57 .0 amps, and we have a final current of zero, and this happens in a change in time of 28 milliseconds, what is going to be the averaged induced emf in our loop? so we know from faraday's law, our induced emf is the opposite of our change in flux over our change in time.

02:28 And we know, well, our flux is going to be changing, but what's changing is going to be b, not a.

02:35 So we're going to have a change in b times our area divided by a change in time.

02:40 Furthermore, we know that b inside our loop here is going to be given by the b of the solenoid.

02:46 That's what's creating it.

02:47 So we're going to have negative area over delta time times b final, which in this case is going to be mu naught n over l times i final, minus b initial, which is mu naught n over l times i initial.

03:02 Well, since we know the final current is going to be zero, this reduces down kind of nice.

03:07 So we're going to have an induced current of positive area, which in this case is going to be two pi times two r squared, without the two there, just pi r squared, right? divided by delta t times mu naught n over l times i naught.

03:25 And we have all of those values, so we can go ahead and plug them in.

03:29 And if i do that, i'm going to get a magnitude of my induced voltage of 0 .101 volts, which is also 101 millivolts...

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