Question Video: Calculating the Number of Turns of a Current-Carrying Solenoid | Nagwa Question Video: Calculating the Number of Turns of a Current-Carrying Solenoid | Nagwa

Question Video: Calculating the Number of Turns of a Current-Carrying Solenoid Physics • Third Year of Secondary School

A wire that carries a constant current of 0.9 A is formed into a solenoid of length 310 mm. The strength of the magnetic field at the center of the solenoid is measured to be 7.7 × 10⁻⁴ T. Calculate the number of turns used to form the solenoid, giving your answer to the nearest whole number of turns. Use a value of 4𝜋 × 10⁻⁷ T⋅m/A for 𝜇₀.

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Video Transcript

A wire that carries a constant current of 0.9 amperes is formed into a solenoid of length 310 millimeters. The strength of the magnetic field at the center of the solenoid is measured to be 7.7 times 10 to the negative fourth teslas. Calculate the number of turns used to form the solenoid, giving your answer to the nearest whole number of turns. Use a value of four 𝜋 times 10 to the negative seventh tesla meters per ampere for 𝜇 naught.

In this example then, we have a solenoid that carries a constant current, we can call it capital 𝐼, of 0.9 amperes and whose length is 310 millimeters. And we’ll call this length capital 𝐿. The solenoid is made of an unknown number of turns. We’ll call that number capital 𝑁, and that’s what we want to solve for. To help us do this along with the current 𝐼 and the length 𝐿, we’re told the strength of the magnetic field at the solenoid center. We can call that field 𝐵, and its strength is given as 7.7 times 10 to the negative fourth teslas. In order to calculate capital 𝑁, the total number of turns in the solenoid, we’ll want to recall how this variable is related to the variables of magnetic field strength, current, and length.

The strength of the magnetic field at the center of a solenoid is given by 𝜇 naught, this constant called the permeability of free space, multiplied by the total number of turns in the solenoid times the current that exists in it all divided by the length of the solenoid along its axis. In our case, it’s not 𝐵 that we want to solve for but the number of turns, 𝑁. So to do that, let’s multiply both sides of the equation by 𝐿 over 𝜇 naught times 𝐼. Over on the right-hand side, this means that 𝐿, 𝜇 naught, and 𝐼 all cancel out. And we find that the number of turns in a solenoid is equal to its length times the magnetic field strength at its center divided by 𝜇 naught times the current in the solenoid 𝐼.

When it comes to the factors on the left-hand side of this expression, we’re given all four of them. We know 𝐿, 𝐵, and 𝐼. And we’re told to use a value of four 𝜋 times 10 to the negative seventh tesla meters per ampere for 𝜇 naught. Substituting all these values in, we find this expression for the number of turns 𝑁. Before we calculate this value though, let’s convert this length of our solenoid, which is currently in units of millimeters, to units of meters. To help us do that, we can recall that 1000 millimeters equals one meter, which means that to convert 310 millimeters to meters, we’ll shift our decimal place one, two, three spots to the left, giving us a result of 0.310 meters.

And now, let’s look at the units in the numerator and denominator of this fraction. We see, first of all, that these units of meters in numerator cancel with meters here. And then also, the unit of teslas cancels out from top and bottom, as does the unit of amperes because the ampere is in the numerator and denominator, we could say, of our overall denominator. So just like we would hope for, this result is going to be unitless because we’re calculating a pure number.

Now, when we enter this fraction on our calculator, we find that we don’t actually get a whole number result. This may happen practically due to the construction of a solenoid where, say, at the ends of the solenoid, one complete turn may not be finished. So, there’s nothing necessarily wrong that 𝑁 is not a whole number. But our question statement does tell us to round our result to the nearest whole number. When we do this, we find a result of 211. This is the number of turns this solenoid has, to the nearest whole number.

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