Video: Calculating the Strength of the Magnetic Field Produced by a Solenoid

A solenoid has 15.0 turns per centimeter. What current will produce a magnetic field of 3.00 Γ— 10⁻² T within the solenoid?

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Video Transcript

A solenoid has 15.0 turns per centimeter. What current will produce a magnetic field of 3.00 times 10 to the negative two tesla within the solenoid?

We can call the number of turns per centimeter, 15.0, lowercase 𝑛. The magnetic field within the solenoid, 3.00 times 10 to the negative two teslas, we’ll name 𝐡. We want to solve for the current that will produce that magnitude of a magnetic field. We’ll call that current 𝐼. We can begin our solution by recalling the mathematical relationship for the magnetic field inside a solenoid.

When a solenoid has an air core, as in this case, then the magnetic field within it is equal to the permeability of free space, πœ‡ naught, times the number of turns of coil per unit length multiplied by the magnitude of the current running through those coils. πœ‡ naught is a constant whose value we’ll assume as exactly 1.26 times 10 to the negative six tesla meters per ampere. So the magnetic field in a solenoid is πœ‡ naught 𝑛 times 𝐼.

Rearranging to solve for 𝐼, we see it equals 𝐡 over πœ‡ naught times 𝑛. The number of turns per centimeter, 𝑛, and the magnetic field strength, 𝐡, are both given to us. And πœ‡ naught is a known constant. So we’re now ready to plug in and solve for 𝐼. When we plug in for these values, we make sure to convert our number of turns per unit length to number of turns per meter so that the units are consistent with the units in the rest of this expression.

When we enter these values on our calculator, we find that 𝐼, to three significant figures, is 15.9 amps. That’s the current running through the loops of this solenoid.

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