The magnetic field inside an air-filled solenoid 38.0 cm long ...

Question

The magnetic field inside an air-filled solenoid 38.0 cm long and 2.10 cm in diameter is 0.720 T. Approximately how much energy is stored in this field?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. What is the energy stored in the magnetic field of a coil used in a physics lab experiment where the coil has a length of 50.0 centimeters, a diameter of 3.00 centimeters and generates a magnetic field of 0.500 teslas. OK. So it appears the final answer that we're ultimately trying to solve for our end goal of sorts is we're trying to figure out what the energy stored in this magnetic field of a coil that's used in this physics lab experiment provided the conditions set by the prom itself. So now that we know that we're trying to figure out what the energy stored is inside of this magnetic field inside of the coil. Let's read off our multiple choice answers to see what our final answer for the energy might be. And let us note that they're all in the same units of jewels. So a is 22 B is 35 C is 55 and D is 87. OK. So first off, let us write down all of our known variables. So we know that the magnetic field, which we're gonna call B is equal to 0.500 teslas. We also know the length of the coil which we're gonna call lowercase L. And the length of the coil L is equal to 50.0 centimeters, which is also equal to 0.500 m, which I've gone ahead and done this conversion from centimeters to meters for you to save time. But you can use dimensional analysis or you could just quickly look it up. We also know the diameter of the coil which we're gonna call lowercase D and the diameter of the coil is equal to 3.00 centimeters or it is also equal to 0.0300 m. Awesome. We also know the radius of the coil because as we should recall, the radius R is equal to the diameter divided by two, which is equal to in this case, since our diameter is 0.0300 m divided by two. That would mean when we provide that or when we put that, that equation into our calculator, we will know that our final answer for the radius is 0.0150 m. Fantastic. And we also know it doesn't directly state the problem itself doesn't directly state this. But we also know the permeability of free space D Tron call mu zero. Or sometimes you might hear it as mu not but mu zero, which is the permeability of free space. This constant is equal to four multiplied by pi multiplied by 10 to the power of negative seven and its units are teslas multiplied by meters per amp fantastic. So now at this stage, we need to identify and use the relevant formula that will help us solve for our final answer, which ultimately we're trying to figure out what the energy stored in the magnetic field of the coil is. So in order to do that, let us recall and use the equation that helps us solve for the energy stored in the magnetic field of a coil which states that the energy E is equal to one half multiplied by V squared, which is the magnetic field squared divided by zero, which is the permeability of free space multiplied by V the volume. So now at this stage, let's take a moment here off to the side and let us recall and use the equation to help us sulfur the volume of a coil. So let us recall that the volume of a coil is equal to. So V the volume is equal to pi multiplied by the radius squared multiplied by the length of the coil. Awesome. So now at this stage, we need to substitute in all of our known variables and solve for the energy, which is our final answer that we're ultimately trying to solve for. So when we plug in our known variables, we'll find out that the energy is equal to one half multiplied by our magnetic field, which is 0.500 teslas square divided by the permeability of free space, which once again is four multiplied by pi multiplied by 10 to the power of negative seven teslas multiplied by meters per amp awesome. And then we need to multiply this by the volume which states that the volume V is equal to pi multiplied by our radius, which our radius is 0.0150. And don't forget that its units are meters squared multiplied by the length which is 0.500 meters. And don't forget your other bracket. So when we plug that expression into our calculator, we will find that E the energy when rounding to the nearest whole number or to two significant figures, the energy E is approximately equal to 35 joules and that's it. That's our final answer. Hooray, we did it. So the energy stored in the magnetic field of the coil is approximately 35 jewels and this corresponds with multiple choice answer letter B which once again is 35 jewels. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

The magnetic field inside an air-filled solenoid 38.0 cm long and 2.10 cm in diameter is 0.720 T. Approximately how much energy is stored in this field?

Key Concepts

Magnetic Field in a Solenoid

Energy Density of a Magnetic Field

Volume of the Solenoid

Watch next