A small solenoid (radius rₐ ) is inside a larger solenoid (rad...

Question

A small solenoid (radius rₐ ) is inside a larger solenoid (radius r₆ > rₐ ). They are coaxial with nₐ and n₆ turns per unit length, respectively. The solenoids carry the same current, but in opposite directions. Let r be the radial distance from the common axis of the solenoids. If the magnetic field inside the inner solenoid (r < rₐ) is to be in the opposite direction as the field between the solenoids (rₐ < r < r₆) , but have half the magnitude, determine the required ratio n₆/nₐ .

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with magnetic fields. This problem says in a lab experiment testing magnetic fields, a large solenoid with radius R two surrounds a smaller solenoid with radius R one where R one is less than R two, the solenoids are coaxial with the inner solenoid having a turn density of N one and the outer solenoid having N two both carry the same current I but in opposite directions. The magnetic field inside the smaller solenoid where R is gonna be less than R one is observed to be in the opposite direction and half the magnitude of the field between the solenoids where R one is less than R less, which is less than R two. Based on this, determine the ratio of N two divided by N one to match the measured field. We're given four possible choices as our answers. For choice A we have one divided by two. For choice B, we have two divided by three. For choice C, we have one and for choice D, we have three divided by two. Now we're dealing with solenoids in this problem. So we need to recall our formula for the magnetic field of a solenoid. So recall that that is B is equal to Muno multiplied by I multiplied by N or B. Here is the strength of magnetic field. Muno is the permeability of free space. I is the current and N is the turn density. Now, in the problem, we were given information of a relationship between the magnetic field strength inside the smaller solenoid to the magnetic field strength in between the solenoids. We're gonna actually use this to calculate our ratio of N two divided by N one. So if I look at the magnetic field inside the smaller solenoid, that's gonna actually be um the combined strength of the magnetic fields from the outer solenoid and the inner solenoid. So let's call it the outer solenoids magnetic field strength B two. And I'm going to have to subtract from that the strength of the magnetic field due to the inner solenoid, which we call B one. The reason why we're subtracting them is because the currents are going in opposite directions. So if the currents are going in opposite directions, that means the direction of the magnetic fields will also be going in opposite directions. So therefore, we're going to subtract them. And here we've just chosen that um the direction of the outer solenoids magnetic field is just gonna be the positive direction. And we wanted to relate this to the magnetic field strength in between the two solenoids and we're told that it is in the opposite direction and half the magnitude. And so we'll have B two minus B one, it's gonna be equal to minus one half. So the minus sign is the opposite direction and the half is due to the half the magnitude. And in between the two solenoids, we only have the magnetic field due to the outer solenoid, which is B two. So we'll take that minus one half and multiply it by B two. And that's because outside of a solenoid, the magnetic field strength is zero. So the inner solenoid doesn't contribute anything to the magnetic field in between the two solenoids. So at this point, we can plug in our expression for the magnetic field for each one of these um terms here. So for B two, we'll have Muno multiplied by I multiplied by N two. That's B minus. For B one, we'll have Muno multiplied by I multiplied by N one. And that's gonna be equal to minus one half multiplied by Muno multiplied by I multiplied by N two. So here we have the same eyes because they have the same current, even though it's going in opposite directions, we've already taken into account the direction of the currents. So if I look at our expression here, the mu knots will cancel out and the eyes will cancel out as well. And what I'm left with is N two minus N one is equal to minus one half multiplied by N two. So I need to solve for this ratio of in one divided by N two. So I'll just collect like terms on each side. So I'll gather the N twos on the left hand side of the equation. And when I do that, I'll get three divided by two, multiplied by N two. And that's gonna be equal to and one. Now, I just need to divide both sides by N one and divide over the three divided by two. And then when I do that, I'll get N two divided by N one is equal to two divided by three. So that is the ratio that I'm looking for. And that actually corresponds to answer choice B. So the trick to setting up this problem relied on two properties of solenoids. One that the direction of the current determines the direction of the magnetic field and two that the magnetic field strength outside of a solenoid is zero. So we use those properties and our equation for the strength of the magnetic field in terms of the current and the current density. And we use that along with the information that we given the problem relating the magnetic field strength inside the smaller solenoid to the magnetic field strength in between the two solenoids to set up an equation that allowed us to eventually calculate our um ratio of our current densities. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Magnetic Field in Solenoids

Superposition of Magnetic Fields

Turn Density Ratio

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