Magnetic force acting on a current-carrying conductor
If a magnetic force is exerted on a single charged particle when the particle moves through a magnetic field, it should not surprise you that a current-carrying wire also experiences a force when placed in a magnetic field. This follows from the fact that the current is a collection of many charged particles in motion; hence, the resultant force exerted by the field on the wire is the vector sum of the individual forces exerted on all the charged particles making up the current. The force exerted on the particles is transmitted to the wire when the particles collide with the atoms making up the wire.
To indicate the direction of B in illustrations, we sometimes present perspective views, such as those in Figures 29.5, 29.6a, and 29.7. In flat illustrations, such as in Figure 29.6b to d, we depict a magnetic field directed into the page with blue crosses, which represent the tails of arrows shot perpendicularly and away from you. In this case, we call the field Bin, where the subscript “in” indicates “into the page.” If B is perpendicular and directed out of the page, we use a series of blue dots, which represent the tips of arrows coming toward you (see Fig. P29.56). In this case, we call the field Bout. If B lies in the plane of the page, we use a series of blue field lines with arrowheads, as shown in Figure 29.8.
One can demonstrate the magnetic force acting on a current-carrying conductor by hanging a wire between the poles of a magnet, as shown in Figure 29.6a. For ease in visualization, part of the horseshoe magnet in part (a) is removed to show the end face of the south pole in parts (b), (c), and (d) of Figure 29.6. The magnetic field is directed into the page and covers the region within the shaded circles. When the current in the wire is zero, the wire remains vertical, as shown in Figure 29.6b. However, when a current directed upward flows in the wire, as shown in Figure 29.6c, the wire deflects to the left. If we reverse the current, as shown in Figure 29.6d, the wire deflects to the right.
Let us quantify this discussion by considering a straight segment of wire of length L and cross-sectional area A, carrying a current I in a uniform magnetic field B, as shown in Figure 29.7. The magnetic force exerted on a charge q moving with a drift velocity vd is qvd x B. To find the total force acting on the wire, we multiply the force qvd x B exerted on one charge by the number of charges in the segment. Because the volume of the segment is AL, the number of charges in the segment is nAL, where n is the number of charges per unit volume. Hence, the total magnetic force on the wire of length L is
We can write this expression in a more convenient form by noting that, from Equation 27.4, the current in the wire is I = nqvdA. Therefore,
where L is a vector that points in the direction of the current I and has a magnitude equal to the length L of the segment. Note that this expression applies only to a straight segment of wire in a uniform magnetic field.
Now let us consider an arbitrarily shaped wire segment of uniform cross-section in a magnetic field, as shown in Figure 29.8. It follows from Equation 29.3 that the magnetic force exerted on a small segment of vector length ds in the presence of a field B is
where dFB is directed out of the page for the directions assumed in Figure 29.8. We can consider Equation 29.4 as an alternative definition of B. That is, we can define the magnetic field B in terms of a measurable force exerted on a current element, where the force is a maximum when B is perpendicular to the element and zero when B is parallel to the element.
To calculate the total force FB acting on the wire shown in Figure 29.8, we integrate Equation 29.4 over the length of the wire:
where a and b represent the end points of the wire. When this integration is carried out, the magnitude of the magnetic field and the direction the field makes with the vector ds (in other words, with the orientation of the element) may differ at different points.
Now let us consider two special cases involving Equation 29.5. In both cases, the magnetic field is taken to be constant in magnitude and direction.
Case 1 A curved wire carries a current I and is located in a uniform magnetic field B, as shown in Figure 29.9a. Because the field is uniform, we can take B outside the integral in Equation 29.5, and we obtain
But the quantity ∫ba ds represents the vector sum of all the length elements from a to b. From the law of vector addition, the sum equals the vector L', directed from a to b. Therefore, Equation 29.6 reduces to
Case 2 An arbitrarily shaped closed loop carrying a current I is placed in a uniform magnetic field, as shown in Figure 29.9b. We can again express the force acting on the loop in the form of Equation 29.6, but this time we must take the vector sum of the length elements ds over the entire loop:
Because the set of length elements forms a closed polygon, the vector sum must be zero. This follows from the graphical procedure for adding vectors by the polygon method. Because ∮ds = 0, we conclude that FB = 0:
“The net magnetic force acting on any closed current loop in a uniform magnetic field is zero”.
The four wires shown in Figure 29.11 all carry the same current from point A to point B through the same magnetic field. Rank the wires according to the magnitude of the magnetic force exerted on them, from greatest to least.