Video: Magnetic Field Within a Solenoid

The current through a solenoid is 2.0 A. How many turns per centimeter must be wound on the solenoid in order to produce a magnetic field of 2.0 × 10⁻³ T within it?

02:03

Video Transcript

The current through a solenoid is 2.0 amperes. How many turns per centimeter must be wound on the solenoid in order to produce a magnetic field of 2.0 times 10 to the negative third tesla within it?

The current through the solenoid, 2.3 amperes, we can name 𝐼. And the desired magnetic field within the solenoid, 2.0 times 10 to the negative third tesla, we’ll call 𝐵. We want to solve for the number of turns per centimeter that must be wound on the solenoid in order to create this magnetic field. We’ll call that value lowercase 𝑛.

To start on our solution, let’s recall the mathematical relationship for the magnetic field within a solenoid. In general, 𝐵 is equal to the magnetic susceptibility of any material in the solenoid’s core multiplied by the permeability of free space, 𝜇 naught, times the number of turns per unit length of the solenoid multiplied by the current, 𝐼, that runs through its coils. The permeability of free space is a constant whose value we’ll treat as exactly 1.26 times 10 to the negative six tesla meters per amp.

In our case, we can write 𝐵 equals 𝜇 naught 𝑛 times 𝐼 because the magnetic susceptibility of our core of air is one. When we rearrange this equation algebraically to solve for 𝑛, we find it’s equal to the magnetic field inside the solenoid divided by 𝜇 naught multiplied by the current that runs through the solenoid coils. We’ve been given the magnetic field and current in the problem statement and 𝜇 naught is a known constant. So we’re now ready to plug in and solve for 𝑛.

When we do and enter these values on our calculator, rounded to the nearest whole number, we find that 𝑛 is equal to eight. That’s the number of turns per centimeter this solenoid would need to have to produce the desired field.

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