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(a) An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner’s 2.00 T field with his fingers pointing in the direction of the field. Find the average emf induced in his wedding ring, given its diameter is 2.20 cm and assuming it takes 0.250 s to move it into the field. (b) Discuss whether this current would significantly change the temperature of the ring.

a. 3 $\mathrm{mV}$ ,b. temperature increase will be small.

02:27

Charles M.

11:17

Mohammad A.

Physics 102 Electricity and Magnetism

Chapter 23

Electromagnetic Induction, AC Circuits, and Electrical Technologies

Electromagnetic Induction

Cornell University

University of Michigan - Ann Arbor

University of Winnipeg

McMaster University

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Yeah. So we are told that an MRI technician moves his hand from a region of low, very low magnetic field strength into an MRI scanners to Tesla field with his fingers pointing in the direction of the field. Um, and we want to find the average PMF induced in his wedding ring. Um, given its 2.2 centimeter diameter and assuming it takes a quarter of a second to move into the field and whether that, um, well lead to a significant temperature change. Um so mhm. What we're going to eat using here is, um, the relationship between the rate of change of the flux, magnetic flux and the electric motor force. Um, so in this case here, um, we have the magnetic fuck. So they're persons fingers like this, and then the ring is around like this, and then the magnetic field is like this, um, along the finger, which means that the magnetic the angle between, um this magnetic flux is the area, and then then the normal to the area is in this direction, and there was the magnetic field. So this is the ring, and the normal to the ring is the area of the cross section of the ring is along the finger, which is in the same direction as the magnetic field. So this coastline of to to find that being one, because data is zero. Um, so then this the magnetic flux is then just the magnetic field time with the the area. The cross sectional area. Um, that's contained. Okay, so and then if you're a ring, we just have one turn. Um, so that's just one. And then we're given we need to get the area, but were given the diameter and 2.2 centimeters. Seems like a pretty darn big ring. But anyway, um, that's the diameter we were given for the ring. And we have the field and then the time. So we have everything we need, um, to calculate this e m f. Yeah, so it looks like Mr Minus sign in here now. So what we have is, um, the e m f or the back. MF is minus b pi d squared over for Delta T. We could just plug all of those in, um and we wind up with a value of three million volts. Um, is the e. M f in the in the ring, which is going to then induce a current. And that current is gonna then ah, generate some some heat or because of the resistance in the ring. Um, actually, in the next problem they give us, they tell us what the resistance is. Um, and so I'll just say Let's say that the resistance is, you know, 0.1 homes. So obviously, if it's a gold ring, it's going to have, you know, Gold is pretty good conductor, actually, very good conductor. And we can just say that, you know, we know that the resistance is going to be very small. Um, but we also see that, you know, the voltage is very small. And so if we know what the power the power is, um, the voltage or the electro motive force times the current and we know the current is the resistance. Um, time sorry. The current is the voltage over the resistance. So the power the power that's being, um, dissipated in the, uh, or chain converted to heat in the ring is electric motor. Four squared over the resistance. Um, so you might think Well, one of the resistance is small, so that's, you know, this might be big, but, you know, this is also small because this is Miller volts, right? So we square Miller Volt, and we divide by something small. We get roughly one million watts. Um, and then if the person's finger is, you know, under, you know, in there for only for a quarter of a second, then we get roughly a quarter of a million jewel of energy being dissipated, which, you know, if you think about you know how you know 100 watts. You know, you know, one Miller, What is a very, very small amount of power? Um, so and then, you know, a quarter of a million jewel is extremely small amount of energy. Um, so there's probably gonna be I mean, I would say the change in temperature is going to be, um, completely unnoticeable. Um, to the to the m r. I technician

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