(II) A 15 -loop circular coil 22 cm in diameter lies in the x y plane. The current

step 1 Okay, so during chapter 27, problem 39. So we have a 15 loop circular coil with a diameter of 22

(II) A 15 -loop circular coil 22 cm in diameter lies...

$$ \begin{array}{l}{\text { (II) A } 15 \text { -loop circular coil } 22 \mathrm{cm} \text { in diameter lies in the } x y} \\ {\text { plane. The current in each loop of the coil is } 7.6 \mathrm{A} \text { clockwise, }} \\ {\text { and an external magnetic field } \vec{\mathbf{B}}=(0.5 \hat{\mathbf{i}}+0.60 \hat{\mathbf{j}}-0.65 \hat{\mathbf{j}}) \mathrm{T}} \\ {\text { passes through the coil. Determine }(a) \text { the magnetic }}\end{array} $$ $$ \begin{array}{l}{\text { moment of the coil, } \vec{\boldsymbol{\mu}} ;(b) \text { the torque on the coil due to the }} \\ {\text { external magnetic field; }(c) \text { the potential energy } U \text { of the }} \\ {\text { coil in the field (take the same zero for } U \text { as we did in our }} \\ {\text { discussion of Fig. } 22}\end{array} $$

$$
\begin{array}{l}{\text { (II) A } 15 \text { -loop circular coil } 22 \mathrm{cm} \text { in diameter lies in the } x y} \\ {\text { plane. The current in each loop of the coil is } 7.6 \mathrm{A} \text { clockwise, }} \\ {\text { and an external magnetic field } \vec{\mathbf{B}}=(0.5 \hat{\mathbf{i}}+0.60 \hat{\mathbf{j}}-0.65 \hat{\mathbf{j}}) \mathrm{T}} \\ {\text { passes through the coil. Determine }(a) \text { the magnetic }}\end{array}
$$ $$
\begin{array}{l}{\text { moment of the coil, } \vec{\boldsymbol{\mu}} ;(b) \text { the torque on the coil due to the }} \\ {\text { external magnetic field; }(c) \text { the potential energy } U \text { of the }} \\ {\text { coil in the field (take the same zero for } U \text { as we did in our }} \\ {\text { discussion of Fig. } 22}\end{array}
$$

Key Concepts

Area of a Circular Loop

This is a basic geometric concept where the area of a circular loop is determined by the formula A = ?r², with r being the radius of the circle. This area plays a direct role in computing the magnetic dipole moment for a current loop since it factors into the product with the current.

Magnetic Potential Energy

The potential energy of a magnetic dipole in an external magnetic field is given by the negative dot product of the magnetic moment and the magnetic field (U = -? · B). This energy reflects the work required to rotate the dipole from a reference orientation to its current orientation in the magnetic field.

Magnetic Dipole Moment

This concept refers to the magnetic strength and orientation of a system that can be represented as a magnetic dipole. For a current-carrying loop, the magnetic dipole moment is defined as the product of the current and the area of the loop, with a direction determined by the right-hand rule. In more complex systems, multiple loops or coils sum their moments vectorially.

Torque on a Magnetic Dipole

A magnetic dipole placed in an external magnetic field experiences a torque that tends to align it with the field. The torque is calculated as the cross product of the magnetic moment and the magnetic field (? = ? × B), which quantifies the rotational effect exerted by the field on the dipole.

Transcript

00:01 Okay, so during chapter 27, problem 39.

00:04 So we have a 15 loop circular coil with a diameter of 22 centimeters, and it lies in the xy plane.

00:14 It says the current in each loop is 7 .6 amps, and it is clockwise.

00:21 An external magnetic field b equaling this 0 .55 in the i direction plus 0 .6 in the j direction and minus 0 .65 in the k direction all teslas passes to the coil so part a says determine the magnetic moment of the coil so mu okay okay so the magnetic moment of the coil so mu okay so the magnetic moment of the coil is given by equation 2710 if you want to go back and look at that and since the current flows in the clockwise direction the right -hand rule shows us that the direction of the magnetic moment must be down which means or sorry down doesn't make much sense but if the current is an x the xy plane down corresponds to the negative z axis or negative k hat direction okay so now the equation tells us that mu is given by n i a hat or a vector so this is 15 loops times the current for 7 .6 and now times our area which is pi times radius 0 .11 meters squared and we already know that this is going to be in the negative k hat direction okay so if we plug all this in and that's that units we get an answer of negative 4 .3 in the k -hat direction and this is amps meter squared.

02:09 This is our u vector for the coil or mu vector for the coil.

02:16 So part b says we use equation we're going to use the equation 2711 because this equation is going to help us find the torque because that's what we want to find the torque on the coil due to the magnetic field.

02:34 So the torque is given.

02:36 By mu cross b so if we want to do this cross product we can take the determinant of the i jk and now we take the mu version which is 0 -0 -0 -43 ams meter squared and now for v it has 0 .55 tesla 0 .6 tesla and negative 0 .65 tesla.

03:09 So we can take the determinant of this and that's the result of a cross product.

03:14 So for this one we have zero, sorry, for the ith, let me take that back.

03:24 For the ith direction, we have zero minus this value.

03:29 So that becomes positive 0 .6 times 4 .3.

03:35 Okay.

03:36 Now for the j, we can block.

03:37 These out, take the determinant this is 0 here, again minus this, but since we're doing the j hat is another negative sign.

03:46 So this is negative 4 .3 times 0 .5.

03:53 And then lastly, we can block out this guy and we see that this is 0 .0.

04:01 So this is in the ith direction and this is in the jth direction and we said there's zero in the kth direction...