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# 物理代写|电磁学代写Electromagnetism代考|Phys132 Torque on a Current Loop in a Uniform Magnetic Field

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## 物理代写|电磁学代写Electromagnetism代考|Torque on a Current Loop in a Uniform Magnetic Field

We showed that a force acts on a current-carrying conductor placed in a magnetic field, given mathematically by Eq. (6.10). Now, consider a current loop placed in a magnetic field, as shown in Fig. 6.8. Equation (6.10) implied that the net magnetic force acting on the loop is zero (see also Eq. (6.14)). However, we will prove in the following that the magnetic field applies a torque on the current loop. For that, consider a rectangular loop carrying a current $I$ in the presence of a uniform magnetic field. First, we assume that the magnetic field is directed parallel to the plane of the loop, as shown in Fig. $6.8$.

The magnetic forces exerted on sides 12 and 34 are both zeroes because these wire segments are parallel to the field $\mathbf{B}$; hence, $\mathbf{L} \times \mathbf{B}=0$ for those two segments. However, the magnetic field is normal to the portions 23 and 34 ; therefore, the magnetic forces on Sects. 23 and 41 are different from zero.

We calculate the magnitude of the forces acting on segments 23 and 41 carrying the current $I$ and length $L=a$ from Eq. (6.8), as
$$F_{23}=F_{41}=\mid I(\mathbf{L} \times \mathbf{B} \mid=I a B$$

However, the direction of $\mathbf{F}{41}$, the force exerted on segment 41, is out of the page, and that of $\mathbf{F}{23}$, the force exerted on segment 23 , is into the page, as indicated in Fig. 6.9. Therefore, those two forces point in opposite directions, and they are not directed along the same line of action; that is, the distance between their lines of action is $b$. Thus, those two forces exert a torque on the loop, such that the loop rotates counterclockwise about the axis passing through the point $O$, as shown in Fig. 6.9. The magnitude of that torque $\tau$ is
$$\tau=F_{23} \frac{b}{2}+F_{41} \frac{b}{2}=(I a B) \frac{b}{2}+(I a B) \frac{b}{2}=I a b B$$
where $b / 2$ is the moment arm about $O$ of each force. Denoting $A=a b$ the area of closed loop, then
$$\tau=I A B$$
We can generalize that result by considering the same current-carrying closed loop, but in a uniform magnetic field making an angle $\theta<90^{\circ}$ with the normal vector to the plane of the loop, as indicated in Fig. 6.10. We also assume that $\mathbf{B}$ is perpendicular to the segment lines 12 and 34 .

## 物理代写|电磁学代写Electromagnetism代考|Motion of a Charged Particle in a Uniform Magnetic Field

From Eq. (6.1), the force exerted on a charged particle moving in a magnetic field is normal to the velocity of the particle, and hence, perpendicular to its displacement. Therefore, the work done on the particle by the magnetic force is zero because the force is perpendicular to the displacement vector of the particle.

As a particular case, we consider a positively charged particle moving in a uniform magnetic field with an initial velocity vector of the particle perpendicular to the field (see Fig. 6.13). The direction of the magnetic field is pointing into the page. Using the right-hand rule in Eq. (6.1), we find that the direction of the magnetic force is pointing toward a single point at the center of a circle. Therefore, the particle is going to move in a circle in a plane perpendicular to the magnetic field.

The particle moves in this way because the magnetic force $\mathbf{F}B$ is perpendicular to both $\mathbf{v}$ and $\mathbf{B}$ and has a constant magnitude of $q v B$. As the force deflects the particle, the directions of both $\mathbf{v}$ and $\mathbf{F}_B$ change continuously, and $\mathbf{F}_B$ points toward the center of the circle at each position of the particle. Thus, force changes only the direction of $\mathbf{v}$, but it does change its magnitude. The rotation is counterclockwise for that positive charge, and if $q$ is negative, the rotation would be clockwise. Using the second law of Newton for circular motion, we get $$\sum_i F{i r}=m a_r$$
or
$$F_B=q v B=m \frac{v^2}{r}$$
The radius of the circle is
$$r=\frac{m v^2}{q v B}=\frac{m v}{q B}$$

## 物理代写|电磁学代写Electromagnetism代考|Torque on a Current Loop in a Uniform Magnetic Field

$$F_{23}=F_{41}=\mid I(\mathbf{L} \times \mathbf{B} \mid=I a B$$

$$\tau=F_{23} \frac{b}{2}+F_{41} \frac{b}{2}=(I a B) \frac{b}{2}+(I a B) \frac{b}{2}=I a b B$$

$$\tau=I A B$$

## 物理代写|电磁学代写Electromagnetism代考|Motion of a Charged Particle in a Uniform Magnetic Field

$$\sum_i F i r=m a_r$$

$$F_B=q v B=m \frac{v^2}{r}$$

$$r=\frac{m v^2}{q v B}=\frac{m v}{q B}$$

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