# Question Video: Calculating Current in a Rectangular Loop of Wire in a Uniform Magnetic Field Physics

The diagram shows a rectangular conducting coil with 5 turns that is in a magnetic field with a strength of 650 mT. The sides of the loop parallel to line 𝑑₁ are parallel to the magnetic field, and the sides of the loop parallel to line 𝑑₂ are perpendicular to the magnetic field. The length of 𝑑₁ = 0.055 m, and the length of 𝑑₂ = 0.035 m. The torque on the loop is 1.2 mN⋅m. What is the current in the loop? Answer to the nearest milliampere.

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### Video Transcript

The diagram shows a rectangular conducting coil with five turns that is in a magnetic field with a strength of 650 milliteslas. The sides of the loop parallel to line 𝑑 one are parallel to the magnetic field, and the sides of the loop parallel to line 𝑑 two are perpendicular to the magnetic field. The length of 𝑑 one equals 0.055 meters, and the length of 𝑑 two equals 0.035 meters. The torque on the loop is 1.2 millinewton-meters. What is the current in the loop? Answer to the nearest milliampere.

As we get started with this example, let’s write down some of the information we’re given. We’re told that a rectangular conducting coil here has five turns. We’ll call that value capital 𝑁. We also know that the magnetic field between the poles of this permanent magnet in which the coil exists, we’ll call it 𝐵, is 650 milliteslas. The dimensions of our rectangular coil are 𝑑 one and 𝑑 two. We’ll record those values at the bottom of our page, and lastly, we’re told about the torque that acts on our current-carrying loop. We’ll call that torque 𝜏, and it’s given as 1.2 millinewton-meters.

Whenever we see this, we’ll have to be careful to remember that this first m stands for milli-, the prefix, and the second m stands for meters, the unit. Anyway, knowing all this, we want to solve for the current in the loop. To get started, let’s clear some space on screen. And let’s recall that, in general, the torque that acts on a current-carrying loop of wire in a magnetic field equals the magnetic field strength 𝐵 times the current in the wire 𝐼 multiplied by the cross-sectional area of the loop times the number of turns in the loop all multiplied by the sin of the angle 𝜃.

To understand 𝜃, say that we have our current-carrying loop in our uniform magnetic field 𝐵. If we draw a vector that is perpendicular to the plane of this loop, then the angle between that vector and the external magnetic field 𝐵 is 𝜃. For the current-carrying loop we have here, a vector that is normal or perpendicular to its surface points directly upward. The angle between this vector and the magnetic field is 90 degrees. Therefore, when we apply this relationship, we’ll find that the sin of 𝜃, the sin of 90 degrees, is one. Knowing this, we can write this simplified equation for our expression for torque.

We recall though that it’s not the torque we want to solve for but rather the current in our conductor. If we divide both sides of this equation by 𝐵 times 𝐴 times 𝑁, then those factors all cancel out on the right. We then have an expression where 𝐼, the current, is the subject. Considering the variables on the right-hand side of this expression, we know the torque 𝜏. That’s given to us in the problem statement. We also know the magnetic field 𝐵 and the number of turns in our coil 𝑁. Concerning the area of our loop 𝐴, we’re not directly told this value, but we do know the perpendicular dimensions of the loop, 𝑑 one and 𝑑 two. The product of these two dimensions equals the area 𝐴.

If we substitute in for 𝜏, 𝐵, 𝑑 one, 𝑑 two, and 𝑁, we’re just about ready to calculate and solve for the current 𝐼. Before we do though, we’d like to change the units of our magnetic field from milliteslas to teslas. 1000 milliteslas equals one tesla. So to convert between these two units, we can shift the decimal place one, two, three spots to the left. 650 milliteslas is 0.650 teslas. Now, we could do something similar in our numerator converting millinewton-meters to newton-meters. But in our question statement, we were asked to solve for 𝐼 and round it to the nearest milliampere.

That means if we keep this one milli- prefix on the right-hand side of our expression, then when we calculate this whole fraction, we’ll get an answer in milliamperes. That brings us closer to our final answer, so we’ll take that approach. This fraction is equal to 191.80 and so on milliamperes. And when we round this to the nearest milliampere, eight, being greater than or equal to five, leads us to round up so that our final answer is 192 milliamperes. This is the magnitude of the current that exists in our conductor.

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