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Problem 64 Hard Difficulty

An aluminum ring of radius 5.00 $\mathrm{cm}$ and resistance $3.00 \times$ $10^{-4} \Omega$ is placed around the top of a long air-core solenoid with 1000 turns per meter and a smaller radius of $3.00 \mathrm{cm},$ as in Figure $\mathrm{P} 20.64 .$ If the current in the solenoid is increasing at a constant rate of $270 . \mathrm{A} / \mathrm{s}$ , what is the induced current in the ring? Assume the magnetic field produced by the solenoid over the area at the end of the solenoid is one-half as strong as the field at the center of the solenoid. Assume also the solenoid produces a negligible field outside its cross-sectional area.

Answer

1 \cdot 60 \mathrm{A}

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Video Transcript

So we are asked to find the induced current in the ring where the aluminum ring has a radius which I call our sub are for radius of the ring of 0.5 meters, right and a resistance of capital are of three times 10 to the minus four homes and is placed in a still annoyed, which has 1000 turns per meter. So in his equal, 1000 per meter, Um, in the smaller radius of the solenoid, ourselves s 0.3 centimeters. Okay, um and also tells us the current and the solenoid is increasing at a constant rate. So delta I delta t is 270 amperes per second. So how can we go ahead and find the current than from this well, from owns lock current, I is equal to the induced Ian meth Absalon over Resistance are okay. We only really care about the magnitude of the induced e m f. So let's go ahead and do magnitude of induced ium f here. Okay, so what is the magnitude of the induced Ian met? Because we know that resistance capital are so let's find that magnitude of the induced e M f well, the induce TMS magnitude is equal to the magnitude of the change of magnetic clucks belt If I divided by Delta t. Okay, so this is going to be equal to the magnitude of Delta. Um, what is the flux? Well, it's the magnetic field times the area, the magnitude of that value divided by Delta t Well, the area's not changing, right, But the magnetic field is so this is going to be equal to the area times the change of magnetic field dealt to be divided by Delta t. So then the question is, what's the change of magnetic field? Okay, let's go ahead and get a new page going here. So assuming that the magnetic field produced by the soul annoyed over the area at the end of the solenoid is 1/2 assed strong as the magnetic field at the center of the sill, Annoyed then, ah, dealt to be here. Let's go ahead. Don't to be is going to be equal to 1/2 the area or excuse me, the magnetic field in the solenoid call that Delta bi So, Bess. Okay, well, this is equal to 1/2 you not which is magnetic per me, a bit permeability of free space. It's four times 10 to the minus seven. You can look that up if you want to know more about that constant. It's a very commonly used, constant times the number of turns in the solenoid per meter times the change in current in this Illinois will call that Delta II. Okay, so now plugging these values back in to the expression for the induce TMS magnitude we find that the induced E m f is equal to you. Not so I'm pulling all the constants outfront times, end times the area of the solenoid, which is pie. Times are so this squared all divided by two multiplied by built I over delta t So I just plugged that value in for the change of magnetic field right into the equation. Well, we know everything in this in this expression, you know, it's just that constant. Four pi times 10 to the minus seven. We know the number of turns per meter pies, a constant we know are so best the radius of the solenoid and we know the change in current with respect to time, we were told that the change in car with respect to time is 270 peers per second. So plugging those values in this expression, we find that the magnitude of the induced IMF is 4.79 times 10 to the minus four. And of course the units are volts. So now going back to homes lock where the current is equal to magnitude of the induced E M f, which we just found divided by the resistance, which we were told it's a plugging those values into this expression. We find that this is equal to about 1.6 in the units for current are in Pierre's. That could be boxed in is the solution to our question.

University of Kansas
Top Physics 102 Electricity and Magnetism Educators
Christina K.

Rutgers, The State University of New Jersey

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University of Michigan - Ann Arbor

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University of Washington

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