Skip to main content
Ch 28: Sources of Magnetic Field

Chapter 28, Problem 28

A solenoid is designed to produce a magnetic field of 0.0270 T at its center. It has radius 1.40 cm and length 40.0 cm, and the wire can carry a maximum current of 12.0 A. (b) What total length of wire is required?

Verified Solution
Video duration:
5m
This video solution was recommended by our tutors as helpful for the problem above.
1675
views
1
rank
Was this helpful?

Video transcript

Welcome back everybody. We are tasked with manufacturing a solenoid. With a couple of conditions. We are told that it must generate a magnetic field of 0.15 Zero Tesla's we're told also that it must have a diameter of two cm or .02 m. We're told that it must have a minimum length of centimeters or 500.5 m. And that a current of 60 amps will be able to run through it. Now we are tasked with finding what the total length of the conductor wire will be needed to generate this magnetic field. Lucky, luckily we have a formula for this. We have that the length is going to be equal to the number of turns, times two pi times the radius of each turn. But we have the radius just by dividing the diameter in half. But we need to figure out what this value is right here. So what we can do is look at the formula for the magnitude of a magnetic field. We have another magnetic field is equal to mu not times our N over L times our current. Now this L just to to be very clear here is our minimum length that we need. So let's go ahead and do this. I am going to divide both sides by mu not and I mu not and I and then we get that N over L. S. Is equal to our magnetic field divided by mu not times our current. So let's go ahead and calculate this then. So we have N over L. S. Is equal to 0.15 divided by four pi. Times 10 to the negative, seventh times 60 amps. This gives us 190 now remember however we need this end so we gotta multiply both sides by this L. S. So L. S. This cancels out on this side and we get on the right side times L. S. Which is just the same thing as 1990 times 0.5. Giving us an end value of 995 turns great. Now that we have found that we are ready to find the total length of our conducting wire needed to generate the solenoid and the magnetic field we have L. Is equal to using this formula up here. 995 times two pi times our diameter divided by two. Which will give us the radius. And when you plug this into our calculator we get a final answer of m of coil corresponding to answer choice. D Thank you all so much for watching. Hope this video helped. We will see you all in the next one
Related Practice
Textbook Question
A solid conductor with radius a is supported by insulating disks on the axis of a conducting tube with inner radius b and outer radius c (Fig. E28.43). The central conductor and tube carry equal currents I in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. Derive an expression for the magnitude of the magnetic field (b) at points outside the tube (r > c).
905
views
1
rank
Textbook Question
A 15.0-cm-long solenoid with radius 0.750 cm is closely wound with 600 turns of wire. The current in the windings is 8.00 A. Compute the magnetic field at a point near the center of the solenoid.
950
views
Textbook Question
A solenoid is designed to produce a magnetic field of 0.0270 T at its center. It has radius 1.40 cm and length 40.0 cm, and the wire can carry a maximum current of 12.0 A. (a) What minimum number of turns per unit length must the solenoid have?
656
views
Textbook Question
Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas and as a percentage of the earth's magnetic field, which is 0.50 G. Is this value cause for worry?
981
views
Textbook Question
Four, long, parallel power lines each carry 100-A currents. A cross-sectional diagram of these lines is a square, 20.0 cm on each side. For each of the three cases shown in Fig. E28.25

, calculate the magnetic field at the center of the square.
1089
views
2
rank
Textbook Question
. Two long, parallel wires are separated by a distance of 0.400 m (Fig. E28.29). The currents I1 and I2 have the directions shown. (a) Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other. Is the force attractive or repulsive?
711
views