Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 4.2.4

Torque on a rectangular loop

A rectangular loop in the $x y$ -plane with sides $a$ and $b$ carrying a current $I$ lies in a uniform magnetic field $B = B_{x} a_{x} + B_{y} a_{y} + B_{z} a_{z}$ . Determine the torque on the loop.

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 6-22.

Try it yourself first.

Strategy

Find the force $I l \times B$ on each straight side, then the torque $r \times F$ about the centre. In a uniform field the forces cancel in pairs, so what is left is a pure couple, the same about any point. Compare with $m \times B$ at the end.

1. Dipole moment

Seen from $+ z$ the circulation $C_{1}$ ( $+ a_{x}$ at $y = + b /2$ ) $\to C_{2}$ ( $- a_{y}$ at $x = + a /2$ ) $\to C_{3} \to C_{4}$ is clockwise, so the right-hand rule gives $a_{n} = - a_{z}$ and

m = I S a_{n} = - I ab a_{z} .

2. Forces

In a uniform field $F = I \oint d l \times B$ vanishes for a closed loop ( $\oint d l = 0$ ): the forces on opposite sides cancel in pairs, and what remains is a pure couple. Sides $C_{1}$ and $C_{3}$ (length $a$ , along $\pm a_{x}$ ) carry $\pm F_{1}$ with

F_{1} = I a a_{x} \times B = I a (B_{y} a_{z} - B_{z} a_{y}),

applied at $y = \pm b /2$ (Figure 4.2.19a). Sides $C_{2}$ and $C_{4}$ (length $b$ , along $\mp a_{y}$ ) carry $\pm F_{2}$ with

F_{2} = - I b a_{y} \times B = - I b (B_{z} a_{x} - B_{x} a_{z}),

applied at $x = \pm a /2$ .

3. Torque about the centre

With $T = r \times F$ , Eq. (4.2.39), for each side,

T_{13} T_{24} = \frac{b}{2} a_{y} \times F_{1} - (- \frac{b}{2} a_{y}) \times F_{1} = b a_{y} \times F_{1} = I ab B_{y} a_{x}, = a a_{x} \times F_{2} = - I ab B_{x} a_{y},

where $a_{y} \times a_{z} = a_{x}$ , $a_{y} \times a_{y} = 0$ , $a_{x} \times a_{x} = 0$ and $a_{x} \times a_{z} = - a_{y}$ were used.

Answer

T = I ab (B_{y} a_{x} - B_{x} a_{y}) = m \times B [N \cdot m]

(a) Forces and torque for \vec B=B_y\vec a_y. — (a) Forces and torque for $B = B_{y} a_{y}$ .

(b) |\vec T| versus the angle \theta_B of \vec B from the z-axis. Problem 4.2.4. (a) The sides C_1 and C_3 feel \pm\vec F_1, a couple about the x-axis. The circulation is clockwise seen from +z, so \vec m=-Iab\,\vec a_z. (b) The torque is IabB\sin\theta_B: largest when \vec B lies in the plane of the loop, zero when it is normal to it. — **Figure 4.2.19.** (b) $∣ T ∣$ versus the angle $θ_{B}$ of $B$ from the $z$ -axis. Problem 4.2.4. (a) The sides $C_{1}$ and $C_{3}$ feel $\pm F_{1}$ , a couple about the $x$ -axis. The circulation is clockwise seen from $+ z$ , so $m = - I ab a_{z}$ . (b) The torque is $I ab B sin θ_{B}$ : largest when $B$ lies in the plane of the loop, zero when it is normal to it.

Check

Directly from Eq. (4.2.38),

m \times B = - I ab a_{z} \times (B_{x} a_{x} + B_{y} a_{y} + B_{z} a_{z}) = - I ab (B_{x} a_{y} - B_{y} a_{x}),

the same. $B_{z}$ produces no torque: it only pulls the four sides outward and tries to stretch the loop. Only the in-plane part counts, so $∣ T ∣ = I ab B_{x}^{2} + B_{y}^{2} = I ab B sin θ_{B}$ , largest when $B$ lies in the plane of the loop and zero when $B$ is normal to it (Figure 4.2.19b). The result $T = m \times B$ holds for a plane loop of any shape in a uniform field: the loop can be cut into thin rectangles whose internal sides cancel.

Problem 4.2.4: Torque on a rectangular loop