Question Video: Determining the Acceleration of a Conducting Rod in a Uniform Magnetic Field | Nagwa Question Video: Determining the Acceleration of a Conducting Rod in a Uniform Magnetic Field | Nagwa

# Question Video: Determining the Acceleration of a Conducting Rod in a Uniform Magnetic Field Physics • Third Year of Secondary School

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The circuit shown in the diagram contains a 4.5 v battery attached to smooth conducting rails. The ends of the conducting rails are attached to a 15 cm long conducting rod with a resistance of 2.5 Ω and a mass of 750 g. The circuit is within a 125 mT uniform magnetic field. What is the magnitude of the acceleration of the rod?

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### Video Transcript

The circuit shown in the diagram contains a 4.5-volt battery attached to smooth conducting rails. The ends of the conducting rails are attached to a 15-centimeter-long conducting rod with a resistance of 2.5 ohms and a mass of 750 grams. The circuit is within a 125-millitesla uniform magnetic field. What is the magnitude of the acceleration of the rod?

Let’s start by recalling that moving charges in a magnetic field experience a magnetic force represented by 𝐹 sub 𝐵. So, because this conducting rod is connected as part of a circuit, and since it’s within a magnetic field, a magnetic force will act on the rod and cause it to accelerate. Our job is to find the magnitude of that acceleration.

To start, we know that force is equal to mass times acceleration or that acceleration equals force divided by mass. And we already know the mass of the rod, so we just need to calculate the magnetic force acting on it.

To do this, let’s recall that for a straight conductor of length 𝐿 that’s carrying a current 𝐼 and moving perpendicular to the direction of a magnetic field of strength 𝐵, the magnetic force on the conductor is given by 𝐵 times 𝐼 times 𝐿. We know values for 𝐵 and 𝐿, but we don’t know the value of the current in the rod, 𝐼. So let’s recall that by Ohm’s law, current in a circuit equals the potential difference across the circuit divided by its resistance. We can go ahead and substitute this expression into the formula for the magnetic force. So it becomes 𝐵 times potential difference over resistance times 𝐿.

Finally, we have an expression for 𝐹 sub 𝐵 in terms of values that we know. But remember, we’re really wanting to solve for acceleration. So let’s divide this entire expression by the mass of the rod. Doing this and moving resistance to the denominator, we have our equation for the rod’s acceleration due to the magnetic force. We’re getting really close. But before we go any further, we should make sure that all the values for the terms on the right-hand side are expressed in their appropriate base SI or SI-derived units.

Let’s start with the strength of the magnetic field, 𝐵, which equals 125 milliteslas. This should be written in plain teslas. So 𝐵 equals 0.125 teslas. Next, the potential difference equals 4.5 volts. So it’s good to go, along with resistance, which equals 2.5 ohms. Moving on, the length of the rod 𝐿 is given as 15 centimeters. So let’s rewrite this as 0.15 meters. And last, we know that the mass of the rod is 750 grams or 0.750 kilograms.

We’re now ready to plug these values in. And because they’re all expressed in their appropriate units, we know that we’ll end up with an acceleration value in proper units of meters per second squared. Finally, grabbing a calculator, we get a result of 0.045 meters per second squared. This is the magnitude of the acceleration of the rod.

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