Question Video: Determining the Number of Loops in a Circular Coil of Wire Using Magnetic Field Strength | Nagwa Question Video: Determining the Number of Loops in a Circular Coil of Wire Using Magnetic Field Strength | Nagwa

Question Video: Determining the Number of Loops in a Circular Coil of Wire Using Magnetic Field Strength Physics

A thin, circular coil of wire with a radius of 22 mm that has 𝑁 turns carries a constant current of 0.45 A. The strength of the magnetic field produced by the current is 2.3 Γ— 10⁻⁴ T at the center of the coil. Calculate 𝑁 to the nearest whole number of turns. Use a value of 4πœ‹ Γ— 10⁻⁷ Tβ‹…m/A for πœ‡β‚€.

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Video Transcript

A thin, circular coil of wire with a radius of 22 millimeters that has 𝑁 turns carries a constant current of 0.45 amps. The strength of the magnetic field produced by the current is 2.3 times 10 to the power of negative four tesla at the center of the coil. Calculate 𝑁 to the nearest whole number of turns. Use a value of four πœ‹ times 10 to the power of negative seven tesla meters per amp for πœ‡ naught.

In this question, we need to work out the number of turns of a circular coil of wire, given the radius of the coil, the current in the wire, and the strength of the magnetic field at its center. To answer this question, we need to recall the formula for the magnetic field at the center of a coil of wire. The magnetic field strength, 𝐡, at the center of a coil of wire with 𝑁 turns is equal to πœ‡ naught 𝑁𝐼 divided by two π‘Ÿ, where 𝐼 is the current in the wire and π‘Ÿ is the radius of the coil. πœ‡ naught is a constant known as the vacuum permeability.

In this question, we’ve been given the magnetic field strength 𝐡, the current 𝐼, the radius π‘Ÿ of the coil, and a value for the vacuum permeability. We need to use this information to find the number of turns in the coil. To do this, we’ll first need to rearrange this equation to make 𝑁 the subject. To do this, we start by multiplying both sides of the equation by two π‘Ÿ. This leaves us with two π‘Ÿπ΅ on the left-hand side of the equation. On the right-hand side, the two π‘Ÿ terms in the numerator and the denominator cancel, leaving us with πœ‡ naught 𝑁𝐼.

To get 𝑁 on its own, we now just divide both sides of the equation by πœ‡ naught times 𝐼. This leaves us with the formula 𝑁 equals two π‘Ÿπ΅ divided by πœ‡ naught 𝐼. Now all we need to do is substitute in the values we’ve been given for these quantities and then calculate the answer.

Before we do this though, we need all the quantities to be in their SI base units. The SI base unit for distance is meters, not millimeters. So, we need to rewrite this as π‘Ÿ is equal to 22 times 10 to the power of negative three meters. If we sub all of these quantities into our equation, we find that 𝑁 is equal to two times 22 times 10 to the power of negative three meters multiplied by 2.3 times 10 to the power of negative four tesla divided by four πœ‹ times 10 to the power of negative seven tesla meters per amp times 0.45 amps.

Before we calculate this value, let’s check what units our answer is going to be in. We can see that in the numerator, we have units of teslas multiplied by meters. In the denominator, we have units of tesla meters per amp multiplied by amps. Here, the ampere terms cancel out, again giving us units of tesla meters. So in our expression for 𝑁, both the numerator and denominator have the same units: tesla meters. Overall, these units cancel, leaving us with no units at all. This is exactly what we’d expect, seeing as 𝑁 is just a number.

Now, if we plug all these values into a calculator, we find that 𝑁 is equal to 17.896. Since we’re asked to give 𝑁 to the nearest whole number, our final answer to this question is 18. There are 18 turns of wire in this coil.

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