Question Video: Finding the Strength of the Magnetic Field Inside of a Solenoid | Nagwa Question Video: Finding the Strength of the Magnetic Field Inside of a Solenoid | Nagwa

Question Video: Finding the Strength of the Magnetic Field Inside of a Solenoid Physics • Third Year of Secondary School

A wire that carries a constant current of 0.15 A is formed into a solenoid with 11 turns per centimeter. Calculate the strength of the magnetic field at the center of the solenoid. Give your answer in teslas expressed in scientific notation to one decimal place. 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 0.15 amperes is formed into a solenoid with 11 turns per centimeter. Calculate the strength of the magnetic field at the center of the solenoid. Give your answer in teslas expressed in scientific notation to one decimal place. Use a value of four 𝜋 times 10 to the negative seven tesla meters per ampere for 𝜇 naught.

This question is asking us about a solenoid, which is a wire that’s shaped into a series of equally spaced loops or turns as shown here. We’re told that the wire carries a constant current of 0.15 amperes, which we’ve labeled as 𝐼. As a result of this current, there’s a magnetic field inside of the solenoid, and the strength of this field, which we’ll label as 𝐵, is exactly what we’re asked to find in this question. We can recall that the strength of the magnetic field inside of a solenoid of total length 𝐿 that consists of 𝑁 turns of wire and carries a current of 𝐼 is equal to a constant 𝜇 naught, the permeability of free space, multiplied by the number of turns, capital 𝑁, multiplied by the current 𝐼 divided by the solenoid’s length 𝐿.

On the right-hand side of the equation, we know the current 𝐼 in the wire and we’re also given a value for the constant 𝜇 naught. However, we don’t know the total number of turns of wire, capital 𝑁, and we don’t know the solenoid’s length 𝐿. What we are told though is that the solenoid has 11 turns of wire per centimeter. If we label the number of turns of wire per unit length of the solenoid as lowercase 𝑛, then we can say that lowercase 𝑛 is equal to 11 centimeters to the negative one. Since the SI base unit of length is not the centimeter but rather the meter, let’s convert this value for lowercase 𝑛 from centimeters to the negative one into meters to the negative one.

In order to do this, let’s recall that one meter is equal to 100 centimeters. If we then divide both sides of this relationship by one meter and by 100 centimeters so that on the left the one-meter terms cancel out, while on the right the 100-centimeter terms cancel, we see that one over 100 centimeters is equal to one over one meter. Since one over units of centimeters is centimeters to the negative one and one over units of meters is meters to the negative one, then we can rewrite this as one over 100 centimeters to the negative one is equal to one meter to the negative one.

Finally, if we multiply both sides of this by 100 so we can cancel the 100s on the left-hand side, we find that one centimeter to the negative one is equal to 100 meters to the negative one. That means that to convert from centimeters to the negative one to meters to the negative one, we multiply by a factor of 100.

Multiplying our value of 11 centimeters to the negative one for lowercase 𝑛 by a factor of 100, we find that lowercase 𝑛 is equal to 11 multiplied by 100 meters to the negative one. This works out as 1100 meters to the negative one, or, in other words, there are 1100 turns of wire per meter of the solenoid’s length. Now, the number of turns of wire per unit length, which is this value for lowercase 𝑛, must be equal to the total number of turns, capital 𝑁, divided by the solenoid’s length 𝐿.

Since in this equation here, we don’t know the value for either capital 𝑁 or for 𝐿, but we do know lowercase 𝑛 and that lowercase 𝑛 is equal to capital 𝑁 divided by 𝐿, let’s use this relationship in order to replace the capital 𝑁 divided by 𝐿 in the equation for the magnetic field strength by lowercase 𝑛, the number of turns per unit length. When we do this, we find that 𝐵, the strength of the magnetic field inside of the solenoid, is equal to the permeability of free space, 𝜇 naught, multiplied by the turns per unit length, lowercase 𝑛, multiplied by the current 𝐼.

Let’s now clear some space on the screen so that we can substitute our values into the right-hand side of this equation. When we substitute in our values for lowercase 𝑛 and 𝐼 along with the value we’re given for the constant 𝜇 naught , we find that 𝐵 is equal to four 𝜋 times 10 to the negative seven tesla meters per ampere, which is our value for the constant 𝜇 naught, multiplied by 1100 meters to the negative one, which is the turns per unit length lowercase 𝑛, multiplied by 0.15 amperes, which is the current 𝐼 in the wire.

If we look at the units on the right-hand side, we can see that the meters and the meters to the negative one cancel each other out, and likewise the amperes and the per ampere also cancel. This just leaves us with teslas as our units for 𝐵, the strength of the magnetic field. Evaluating this expression, we find that 𝐵 is equal to 2.073 et cetera times 10 to the negative four tesla. The question wants our answer in teslas expressed in scientific notation to one decimal place. We have the correct units, and our answer is already in scientific notation, which means we just need to round the value to one decimal place.

Rounding to one decimal place, we get our final answer to the question. The strength of the magnetic field at the center of the solenoid is equal to 2.1 times 10 to the negative four tesla.

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