Irodov 3.269: A Small Loop Feels the Gradient – Amit & Deepali Sinha Foundation

Problem Irodov 3.269

A long straight wire carries a current $I$ . A small loop with magnetic moment $\vect p_m$ sits at a distance $r$ from the wire, $\vect r$ being the radius vector from the wire to the loop. Find the magnitude and direction of the force on the loop when $\vect p_m$

is parallel to the wire;
points along $\vect r$ ;
points along the magnetic field of the current $I$ at the loop.

The long wire with current I, the radius vector r to the small loop, and the loop's moment p_m in the three cases: (a) along the wire, (b) along r, (c) across both, along the wire's field at the loop. — The long wire with current $I$ , the radius vector $\vect r$ to the small loop, and the loop’s moment $\vect p_m$ in the three cases: (a) along the wire, (b) along $\vect r$ , (c) across both, along the wire’s field at the loop.

Given

current $I$
distance $r$
moment $\vect p_m$ , (a) $\parallel$ wire, (b) $\parallel \vect r$ , (c) along the wire’s field at the loop
loop small compared with $r$

Find

the force

\vect F

on the loop in each case

Solution

Step 1The force on a small loop

A small loop of fixed moment $\vect p_m$ in a field $\vect B$ has the energy $-\vect p_m\cdot\vect B$ , so the force on it is the gradient of $\vect p_m\cdot\vect B$ , taken with $\vect p_m$ held fixed:

\vect F = \nabla\left(\vect p_m\cdot\vect B\right) \tag{1}

Step 2The wire’s field near the loop

Take $z$ along the wire in the sense of the current, and put the loop on the $x$ axis at $(r, 0, 0)$ . By symmetry the field circles the wire, in the right-hand sense about the current, with the same magnitude all round a circle of radius $\rho$ , so Ampere’s law gives

B\cdot 2\pi\rho = \mu_0 I

Resolving this field of magnitude $\mu_0 I/2\pi\rho$ at a point $(x, y)$ , where $\rho^2 = x^2 + y^2$ , gives the components

B_x = -\frac{\mu_0 I}{2\pi}\,\frac{y}{x^2 + y^2}, \qquad B_y = \frac{\mu_0 I}{2\pi}\,\frac{x}{x^2 + y^2}, \qquad B_z = 0 \tag{2}

At the loop the field is $B = \mu_0 I/2\pi r$ along $+y$ : in this frame, $\vect r$ is along $x$ , the wire along $z$ , and $\vect B$ along $y$ .

Step 3(a) The moment along the wire

With $\vect p_m$ along $z$ , the product $\vect p_m\cdot\vect B = p_mB_z$ vanishes everywhere by (2), so its gradient (1) does too:

\vect F = 0

(3)

Step 4(b) The moment along the radius vector

(b) Looking down the wire, current out of the page: p_m along r, and the force opposite to B at the loop. — (b) Looking down the wire, current out of the page: $\vect p_m$ along $\vect r$ , and the force opposite to $\vect B$ at the loop.

(c) p_m along B (up the page here): the force points to the wire. — (c) $\vect p_m$ along $\vect B$ (up the page here): the force points to the wire.

With $\vect p_m = p_m\vect{i}$ , the force (1) is $p_m\nabla B_x$ . Differentiating $B_x$ from (2) and evaluating at $(r, 0)$ :

\begin{aligned} \frac{\partial B_x}{\partial x} &= \frac{\mu_0 I}{2\pi}\,\frac{2xy}{(x^2 + y^2)^2} = 0\\[2pt] \frac{\partial B_x}{\partial y} &= -\frac{\mu_0 I}{2\pi}\,\frac{x^2 - y^2}{(x^2 + y^2)^2} = -\frac{\mu_0 I}{2\pi r^2} \end{aligned}

so only a $y$ component survives, against the field’s direction there:

\vect F = -\frac{\mu_0 I p_m}{2\pi r^2}\,\frac{\vect B}{B}

(4)

The force has magnitude $\mu_0 Ip_m/2\pi r^2$ and points opposite to $\vect B$ at the loop, along the circle round the wire.

Step 5(c) The moment along the field

With $\vect p_m = p_m\vect{j}$ , along $\vect B$ at the loop, the force is $p_m\nabla B_y$ . Differentiating $B_y$ from (2) and evaluating at $(r, 0)$ :

\begin{aligned} \frac{\partial B_y}{\partial x} &= \frac{\mu_0 I}{2\pi}\,\frac{y^2 - x^2}{(x^2 + y^2)^2} = -\frac{\mu_0 I}{2\pi r^2}\\[2pt] \frac{\partial B_y}{\partial y} &= -\frac{\mu_0 I}{2\pi}\,\frac{2xy}{(x^2 + y^2)^2} = 0 \end{aligned}

so only an $x$ component survives, toward the wire:

\vect F = -\frac{\mu_0 I p_m}{2\pi r^2}\,\frac{\vect r}{r}

(5)

The force has the same magnitude $\mu_0 Ip_m/2\pi r^2$ and points against $\vect r$ : the loop is pulled toward the wire.

Check

Case (c) follows directly from the energy: on the $x$ axis $\vect p_m\cdot\vect B = p_m\mu_0 I/2\pi r$ , and the radial force is its derivative,

F_r = \frac{d}{dr}\left(\frac{\mu_0 Ip_m}{2\pi r}\right) = -\frac{\mu_0 Ip_m}{2\pi r^2},

as in (5): a moment aligned with the field is drawn toward the stronger field near the wire. In (b) the radial derivative gives nothing, because $\vect B$ has no component along $\vect r$ anywhere on the $x$ axis; the force comes instead from the field turning as the loop moves round the wire. Units: $\mu_0 I p_m/r^2$ is (N/A $^2$ )(A)(A $\cdot$ m $^2$ )/m $^2$ $=$ N.

Remark

The wire’s field is the field of a charged thread turned through a right angle about the line, so these answers mirror those for an electric dipole near a charged thread (3.46): a moment along the line feels nothing, a moment along the field is drawn toward the line, and a moment across the field is pushed sideways with a force of the same size. The force is not along $\vect B$ in general, and in (b) it is not even toward or away from the wire; a magnetic moment simply moves toward stronger field only when it is aligned with the field, as in (c).