A wire along the x-axis carries current I in the negative x-di...

Question

A wire along the x-axis carries current I in the negative x-direction through the magnetic field $\vec{B} = {\begin{matrix} B_{0} \frac{x}{l} \hat{k} & 0 \leq x \leq l \\ 0 & elsewhere \end{matrix}$ . Find an expression for the net torque on the wire about the point x = 0.

Accepted Answer

Welcome back, everyone in this problem, a long current carrying conductor oriented parallel to the Y axis lies in a perpendicular magnetic field given by B equals BNY squared divided by LK for the interval where Y is between zero and L B is zero outside these limits a current I flows to the conductor in a positive Y direction. And we want to derive an equation for a torque on the conductor about the point Y equals zero. For our answer choices. A says it's a third of IB not multiplied by L cubed B says it's a half of IB not L squared. C says it's IB not L squared. And D says it's a quarter of IB not multiplied by L to the power of four. Now, first, let's try to get an understanding of what we're really talking about here. So we're talking about a current carrying conductor that's parallel to the Y axis. So let me try to represent that on a diagram. OK. So we have our Y axis here which in three, when we're talking about vectors, we can refer to it as J OK. And of course, we're gonna have K and we're gonna have I now we know that our magnetic field is perpendicular to our current carrying conductor. OK. And it's represented by an expression, we also know that a current I flows to the conductor in the positive Y direction. So we can represent that here as I and we also know that our magnetic field is acting between the interval where Y equals zero. OK can represent that here and where Y equals L. And we're trying to find an equation for the torque on the conductor. In other words, an equation for the torque on the conductor here that represent that as art to right about the point Y equals zero. So essentially, we're trying to find an expression that relates our magnetic field and our length to the torque on the conductor. Now, first of all, let's ask ourselves, what do we know about torque? Well, recall that torque, OK. Torque equals R multiplied by delta F multiplied by sine theta. OK. Where R is the length of the lever arm delta F is the change in force and theta is our angle. OK. And no, let's consider the wire to be formed of small segments. Since each segment will experience a torque. If we can find an equation to represent the torque for each segment, then we should be able to find a total torque on the wire. And thus a torque expression for the conductor about point Y equals zero. OK. So first let's start by thinking about the information that we have here. How can we relate, talk to our current carrying conductor? Well, since we're talking about each segment, OK, then we know that the length of our liver arm equals Y. So for each segment, then the length of our liver arm will be equal to Y OK. Next, we also, we also know that here the force that we're referring to delta F or change in force must be related to our conductor in a magnetic field. Now, what do we know about a current carrying conductor in a magnetic field? Well, the force that's generated is going to be equal to the current. So not delta F but rather the force that's generated is going to be equal to the current multiplied by the length multiplied by the strength of the magnetic field. Therefore, delta F is going to be equal to our current, which is constant multiplied by delta L. OK, which is our change in distance multiplied by our magnetic field B, which is also constant. Now, we know that here, since we're talking about the interval for Y between zero and L or delta F is really our current multiplied by a magnetic field multiplied by delta Y. OK. And finally, we also know that our magnetic field B is perpendicular to our current carrying conductor. So our angle theta equals 90 degrees. So using that information, then that means the torque, the torque for each segment is going to be equal to the length right segment Y multiplied by delta F which is going to be equal to our current multiplied by our magnetic field. Now, in this case, our magnetic field as a function of the initial magnetic field can be represented as B of Y. OK. And that is going to be multiplied by delta Y. So this is an equation that represents the torque at about the point Y equals zero at each segment. So if that's the case, then that means the total talk the to talk it to our torque is going to be equal to the sum of the torque at each segment. OK, which is gonna be equal to the sum of yi multiplied by B of yi multiplied by delta Y. Now notice here we're talking about the so of each segment and as we said, we consider the wire to be formed of small segments. So if we want to find a total torque, we can integrate. So this implies then that our total torque is going to be the integral of our expression. Nor recall that our magnetic field as a function of the initial magnetic field, we already have an equation that represents the magnetic field. So we can substitute it here and then integrate with respect to why to solve for the total torque thus getting our expression. So now that means our talk, when we substitute that expression into our formula, then it's going to be the integral between zero and L because remember Y was between those boundaries or Y is between those boundaries of I multiplied by our expression for B of Y of I, which is B not Y cubed multiplied by DY. In other words, it's the integral of I be not Y cubed with respect to Y between the bounds of zero. And L. Now, if we take out our constants for current and be not here, then we can rewrite our expression as I be not multiplied by the integral of YC with respect to Y between the boundary of zero and L. And that means no, we can integrate. OK. So if we leave our constant oib, then when we integrate Y cubed with respect to Y, the result is Y to the fourth divided by four. So now we'll have Y to the fourth divided by four. Let me write that properly here. That's why to the fourth divided by four between the Boers of L and zero. Now, that means we're gonna here, if we write that out, that's gonna be IB multiplied by L to the fourth divided by four minus zero to the fourth power divided by four. I know zero to the fourth part is zero. So really our torque is going to be equal to I be multiplied by L to the fourth divided by four. Or, or rather I be not, I should have been writing I be not here, forgive me. But I be not which we can write as 1/4 of I, the fourth of I be not multiplied by L to the fourth pore. Therefore, the torque expression for the conductor about the point Y equals zero is 1/4 of I be not multiplied by L to the fourth or as it's written in our solutions. 1/4 of I multiplied by L to the fourth multiplied by B not when we look back on our answer, answer choices that would have been answer choice. D thanks a lot for watching everyone. I hope this video helped.

A wire along the x-axis carries current I in the negative x-direction through the magnetic field $\vec{B} = {\begin{matrix} B_{0} \frac{x}{l} \hat{k} & 0 \leq x \leq l \\ 0 & elsewhere \end{matrix}$ . Find an expression for the net torque on the wire about the point x = 0.

Key Concepts

Magnetic Field

Torque

Lorentz Force

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