Torque on a Current Loop in a Nonuniform Magnetic Field. In...

Torque on a Current Loop in a Nonuniform Magnetic Field. In Section 27.7 the expression for the torque on a current loop was derived assuming that the magnetic field $\vec { B }$ was uniform. But what if $\vec { B }$ is not uniform? Figure $P 27.85$ shows a square loop of wire that lies in the $x y$ -plane. The loop has corners at square loop of wire that lies in the $x y$ -plane. The loop has corners at $( 0,0 ) , ( 0 , L ) , ( L , 0 ) ,$ and $( L , L )$ and carries a constant current $I$ in the clockwise direction. The magnetic field has no z-component but has both $x$ - and $y$ -components: $\vec { \vec { B } } = \left( B _ { 0 } y / L \right) \hat { \boldsymbol { \imath } } + \left( B _ { 0 } x / L \right) \hat { \boldsymbol { J } } ,$ where $B _ { 0 }$ is a positive constant. (a) Sketch the magnetic field lines in the $x y$ -plane. (b) Find the magnitude and direction of the magnetic force exerted on each of the sides of the loop by integrating Eq. $( 27.20 ) .$ (c) If the loop is free to rotate about the $x$ -axis, find the magnitude

Torque on a Current Loop in a Nonuniform Magnetic Field. In Section 27.7 the expression for the torque on a
current loop was derived assuming that the magnetic field $\vec { B }$ was
uniform. But what if $\vec { B }$ is not uniform? Figure $P 27.85$ shows a
square loop of wire that lies in the $x y$ -plane. The loop has corners at square loop of wire that lies in the $x y$ -plane. The loop has corners at
$( 0,0 ) , ( 0 , L ) , ( L , 0 ) ,$ and $( L , L )$ and carries a constant current $I$ in
the clockwise direction. The magnetic field has no z-component but has both $x$ - and $y$ -components: $\vec { \vec { B } } = \left( B _ { 0 } y / L \right) \hat { \boldsymbol { \imath } } + \left( B _ { 0 } x / L \right) \hat { \boldsymbol { J } } ,$ where $B _ { 0 }$ is a positive constant. (a) Sketch the magnetic field lines in the
$x y$ -plane. (b) Find the magnitude and direction of the magnetic force
exerted on each of the sides of the loop by integrating Eq. $( 27.20 ) .$
(c) If the loop is free to rotate about the $x$ -axis, find the magnitude

Key Concepts

Lorentz Force Law

The Lorentz force law describes the force experienced by a charged particle or a current element moving in a magnetic field. For a current-carrying conductor, the differential force can be expressed as dF = I (dl × B), where I is the current, dl is a differential element of the wire, and B is the magnetic field at the location of the element. This principle is foundational in understanding how electric currents interact with magnetic fields.

Current Loop

A current loop is a closed circuit through which a steady current flows. It is significant because the circulation of the current creates a magnetic dipole, and the loop interacts with external magnetic fields. The geometry of the loop influences the distribution of forces and the resultant torque or net force acting on it.

Magnetic Dipole Moment

The magnetic dipole moment is a vector quantity associated with the current loop, defined as ? = I A, where I is the current and A is the vector area of the loop. This concept is crucial because it provides a simplified way to characterize the loop's overall magnetic effects, such as its interaction with nonuniform magnetic fields, by summarizing the effect of distributed currents.

Torque on a Current Loop

The torque on a current loop in a magnetic field arises when there is an angular misalignment between the loop's magnetic dipole moment and the external magnetic field. In a uniform magnetic field, the torque is given by ? = ? × B, which tends to rotate the loop to align ? with the field. This principle also extends to more complex situations, including nonuniform fields, where the distribution of forces over the loop must be considered.

Nonuniform Magnetic Field

A nonuniform magnetic field is one that varies in magnitude and/or direction with position. In such fields, different sections of a current-carrying wire may experience different magnetic forces, leading to net force as well as torque on the current loop. Analyzing the behavior in nonuniform fields often requires integrating the Lorentz force over the entire loop to account for local variations in the magnetic field.

Transcript

00:01 So for the first part, we are going to use force is equal to i times the cross product of length and magnetic field.

00:13 And this gives the magnitude to be i times l times b, sign phi, where phi is the angle between length and the magnetic field.

00:27 And direction is given by right -hand rule.

00:31 So we are going to apply this equation to each segment and see what we get.

00:42 So f pq is going to be i times l pq.

00:49 So current is basically capital i.

00:55 So let's write capital i in each of the case, times b, sine phi.

01:05 So phi for pq segment you see that i and we are parallel.

01:10 So phi is 0 degree and this gives the force to be 0 on the segment.

01:15 For segment pr, we do the same thing, ipr times the magnetic field times sign.

01:26 So here they are perpendicular, so this is 90 degree.

01:31 This gives i times lpr times the magnetic field and we get this value to be 12 newton.

01:42 And by right -hand rule the direction is going to be into the screen or into the page.

01:52 Similarly, we do the same thing for qr, which is i, l -q -r -p times sine phi.

02:07 This is equal to i -times.

02:11 So l -q -r is basically this distance, which can be found from pythagoras, here up because this is basically 90 degree.

02:21 So lqy will then be equal to square root of the sum of the squares of the other two sides times b times sine phi.

02:39 So if this is phi, this is going to be 90 minus 5.

02:47 So this is basically 90 minus 5.

02:50 So this is 5.

02:52 That's what we have.

02:54 So, okay, let's check again.

02:57 So this is basically phi.

03:01 So this is 180 minus 5.

03:03 This is 180 minus 5.

03:06 So this will give this angle to be 5.

03:11 So if this angle is 5, then sine 5 is going to be the perpendicular over hypertenness.

03:23 So that will be l pq over.

03:31 Okay, just a minute.

03:38 Pause.

03:39 Okay, i think, so let's check this value of phi again.

03:47 So this is not the angle between i and v.

03:51 This is the angle.

03:52 So if this is the angle, then this is the phi that we need.

04:01 So that is equal to lpr over lqr.

04:08 I'll just write that.

04:11 S...

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