If you pass a current through a wire, you can create a magnetic field. In this c

Question

If you pass a current through a wire, you can create a magnetic field. In this class we examine the field from three specific configurations: an infinitely-long straight wire, a compact loop of wire (possibly with multiple turns, a coil), or an infinitely long solenoid (think of an infinitely-long cylinder of loops made by wrapping a single wire). See the text for an extended description of all three geometries.
Here’s the challenge: there are three slightly different formulas used to calculate the magnetic field. Which you use depends on the configuration of the wire. Look in the text, check out the figures, and write down the formula for each of these three situations in your notes. Label the formulas and read carefully to make sure you understand what each variable represents.
Suppose you have an infinitely long wire carrying a current of 17.0 A. What is the strength of the magnetic field a distance 0.0340 m. away from the wire?

You are now studying a length of wire that has been coiled up to create a circular coil of 4 turns (individual loops) all arranged compactly. The loop has radius of 0.0320 m and a current of 55.1 A. is passing through the wire. What is the strength of the magnetic field at the center of the loop?

Lastly, you are handed a long solenoid. The special characteristic of an infinitely-long solenoid is that the magnetic field inside the solenoid is uniform. There are 4.71×103 turns over a length of 0.130 m and a current of 2.26 A flows through the wire. What is the strength of the magnetic field inside the solenoid?

This discussion is closed.

Explanation / Answer

formula 1:

Magnetic field due to a long straight Current carrying wire is B = uoi/2pIR

i is current and R is the distance at which we wish to find magnetic field

so

1. B = (4*3.14 e-7 * 17)/(2*3.14 * 0.034)

B = 1 e-4 Tesla or 100 uT

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formula 2: Magnetic field a t the center of circular loop is B = N uoi/2R

B = (4 * 4*3.14 e-7*55.1 )/(2* 0.032)

B = 4.32 mT

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formula 3: Magnetic field due to solonois is B = uo n i

where n is no. of turns per unitm length = N/l

so

B = 4*3.14 e -7 * 2.26 * 4710/0.130

B = 0.102 Tesla

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