Question Video: Finding the Current in a Solenoid | Nagwa Question Video: Finding the Current in a Solenoid | Nagwa

Question Video: Finding the Current in a Solenoid Physics • Third Year of Secondary School

A solenoid is formed of a length of wire that carries a constant current 𝐼. The solenoid has 430 turns of wire per meter. The magnetic field at the center of the solenoid is measured to be 3.2 Γ— 10⁻³ T. Calculate the current, 𝐼, in amperes. Give your answer to 1 decimal place. Use πœ‡β‚€ = 4πœ‹ Γ— 10⁻⁷ Tβ‹…m/A.

04:22

Video Transcript

A solenoid is formed of a length of wire that carries a constant current 𝐼. The solenoid has 430 turns of wire per meter. The magnetic field at the center of the solenoid is measured to be 3.2 times 10 to the negative three tesla. Calculate the current 𝐼 in amperes. Give your answer to one decimal place. Use πœ‡ naught is equal to four πœ‹ times 10 to the negative seven tesla meters per ampere.

In this question, we’re being asked about a solenoid, which is a wire that’s formed into a series of equally spaced loops or turns like we’ve illustrated here. We’re told that this wire carries a constant current of 𝐼, and this current is what we’re being asked to find. As a result of this current in the wire, there’s a magnetic field inside of the solenoid. We’re told that the strength of this magnetic field, which we’ve labeled as 𝐡, is measured at the center of the solenoid to be equal to 3.2 times 10 to the negative three tesla.

We can recall that for a solenoid that has a total length of 𝐿 and consists of 𝑁 turns of wire which carry a current of 𝐼, the magnetic field 𝐡 inside of that solenoid is equal to πœ‡ naught multiplied by capital 𝑁 multiplied by 𝐼 divided by 𝐿, where πœ‡ naught is a constant known as the permeability of free space. This equation here relates the magnetic field 𝐡 inside of the solenoid, which we know the value of, to the current 𝐼 in the wire, which is what we want to work out.

However, we don’t know the total number of turns of wire in this solenoid capital 𝑁, and we also don’t know the solenoid’s length 𝐿. What we are told is that this solenoid has 430 turns of wire per meter. If we define a quantity lowercase 𝑛 as the number of turns of wire per unit length of the solenoid, then we have for this solenoid that lowercase 𝑛 is equal to 430 meters to the negative one because it has 430 turns of wire per meter of length. We can see that the number of turns of wire per unit length, lowercase 𝑛, must be equal to the total number of turns capital 𝑁 divided by the total length 𝐿.

Since in this equation here we don’t know the value for either capital 𝑁 or for 𝐿 but we do know the value of lowercase 𝑛 and that lowercase 𝑛 is equal to capital 𝑁 divided by 𝐿, let’s use this relationship in order to replace capital 𝑁 divided by 𝐿 in this equation for the magnetic field strength by lowercase 𝑛, the number of turns per unit length. When we do this, we find that 𝐡, the strength of the magnetic field inside of the solenoid, is equal to πœ‡ naught, the permeability of free space, multiplied by lowercase 𝑛, the number of turns of wire per unit length, multiplied by the current 𝐼 in the wire. Now, for this solenoid, we know the values of both 𝐡 and lowercase 𝑛. And we’re also given a value for this constant πœ‡ naught. That means that the only unknown quantity in the equation is the current 𝐼 that we’re trying to work out.

In order to work out this current, we need to make 𝐼 the subject of the equation. To do this, we’ll divide both sides of the equation by πœ‡ naught and 𝑁 so that on the right-hand side the πœ‡ naughts and 𝑁’s cancel from the numerator and denominator. If we then swap over the left- and right-hand sides of the equation, we have that the current 𝐼 is equal to 𝐡 divided by πœ‡ naught and lowercase 𝑛.

We’re now ready to go ahead and substitute in our values for the strength of the magnetic field 𝐡, the number of turns of wire per unit length lowercase 𝑛, and the permeability of free space πœ‡ naught. When we do this, we get that 𝐼 is equal to 3.2 times 10 to the negative three tesla divided by four πœ‹ times 10 to the negative seven tesla meters per ampere and 430 meters to the negative one.

Looking at the units on the right-hand side of the equation, we can see that in the denominator the meters and the meters to the negative one cancel each other out. Then, we can also see that the teslas cancel from the numerator and denominator, and this just leaves us with units of one over amperes in the denominator of the fraction. The overall units we’ll get for our current 𝐼 then will be units of one divided by one over amperes. This is simply equal to units of amperes, which is the current unit that the question asks us to use. Evaluating the expression, we calculate a current 𝐼 of 5.922 et cetera amperes.

We should notice that we’re asked to give this answer to one decimal place. Rounding to one decimal place of precision, the result becomes 5.9 amperes. Our answer then is that the current in the wire is equal to 5.9 amperes.

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