# 5.1 Magnetic Forces and Torques

The electric field E at a point in space has been defined as the electric force F e per unit charge acting on a charged test particle placed at that point. We now define the magnetic flux density B at a point in space in terms of the magnetic force F m that acts on a charged test particle moving with velocity u through that point. Experiments revealed that a particle of charge q moving with velocity u in a magnetic field experiences a magnetic force F m given by

F m=q u×B (N).(5.3)

Accordingly, the strength of B is measured in newtons/(C⋅m/s), which is also called the tesla (T). For a positively charged particle, the direction of F m is that of the cross product u×B, which is perpendicular to the plane containing u and B and governed by the right-hand rule. If q is negative, the direction of F m is reversed (Fig. 5-1). The magnitude of F m is given by

F m=q u B sin θ,(5.4)

where θ is the angle between u and B.

# Figure 5-1 The direction of the magnetic force exerted on a charged particle moving in a magnetic field is (a) perpendicular to both B and u and (b) depends on the charge polarity (positive or negative).

Figure 5 1-a Full Alternative Text
Figure 5 1-b Full Alternative Text

If a charged particle resides in the presence of both an electric field E and a magnetic field B, then the total electromagnetic force acting on it is

The force expressed by Eq. (5.5) also is known as the Lorentz force. Electric and magnetic forces exhibit a number of important differences:

1. Whereas the electric force is always in the direction of the electric field, the magnetic force is always perpendicular to the magnetic field.

2. Whereas the electric force acts on a charged particle whether or not it is moving, the magnetic force acts on it only when it is in motion.

3. Whereas the electric force expends energy in displacing a charged particle, the magnetic force does no work when a particle is displaced.

This last statement requires further elaboration. Because the magnetic force F m is always perpendicular to u, F m⋅u=0. Hence, the work performed when a particle is displaced by a differential distance d l=u d t is

d W=F m⋅d l=(F m⋅u)d t=0.(5.6)

# 5.1.1 Magnetic Force on a Current-Carrying Conductor

A current flowing through a conducting wire consists of charged particles drifting through the material of the wire. Consequently, when a current-carrying wire is placed in a magnetic field, it experiences a force equal to the sum of the magnetic forces acting on the charged particles moving within it. Consider, for example, the arrangement shown in Fig. 5-2 in which a vertical wire oriented along the z direction is placed in a magnetic field B (produced by a magnet) oriented along the −x^ direction (into the page). With no current flowing in the wire, F m=0 and the wire maintains its vertical orientation (Fig. 5-2(a)), but when a current is introduced in the wire, the wire deflects to the left (−y^ direction) if the current direction is upward (+z^ direction) and to the right (+y^ direction) if the current direction is downward (−z^ direction). The directions of these deflections are in accordance with the cross product given by Eq. (5.3).

# Figure 5-2 When a slightly flexible vertical wire is placed in a magnetic field directed into the page (as denoted by the crosses), it is (a) not deflected when the current through it is zero, (b) deflected to the left when I is upward, and (c) deflected to the right when I is downward.

Figure 5 2 Full Alternative Text

To quantify the relationship between F m and the current I flowing in a wire, let us consider a small segment of the wire of cross-sectional area A and differential length d l with the direction of d l denoting the direction of the current. Without loss of generality, we assume that the charge carriers constituting the current I are exclusively electrons, which is always a valid assumption for a good conductor. If the wire contains a free-electron charge density ρ v e=−N e e, where N e is the number of moving electrons per unit volume, then the total amount of moving charge contained in an elemental volume of the wire is

d Q=ρ v e A d l=−N e e A d l,(5.7)

and the corresponding magnetic force acting on d Q in the presence of a magnetic field B is

d F m=d Q u e×B=−N e e A d l u e×B,(5.8a)

where u e is the drift velocity of the electrons. Since the direction of a current is defined as the direction of flow of positive charges, the electron drift velocity u e is parallel to d l, but opposite in direction. Thus, d l u e=−d l u e and Eq. (5.8a) becomes

d F m=N e e A u e d l×B.(5.8b)

From Eqs. (4.11) and (4.12), the current I flowing through a cross-sectional area A due to electrons with density ρ v e=−N e e and moving with velocity −u e is

I=ρ v e(−u e)A=(−N e e)(−u e)A=N e e A u e.

Hence, Eq. (5.8b) may be written in the compact form

d F m=I d l×B (N).(5.9)

For a closed circuit of contour C carrying a current I, the total magnetic force is

If the closed wire shown in Fig. 5-3(a) resides in a uniform external magnetic field B, then B can be taken outside the integral in Eq. (5.10), in which case

F m=I(∮C d l)×B=0.(5.11)

# Figure 5-3 In a uniform magnetic field, (a) the net force on a closed current loop is zero because the integral of the displacement vector d l over a closed contour is zero and (b) the force on a line segment is proportional to the vector between the end point (F m=I ℓ×B).

Figure 5 3-a Full Alternative Text

This does not mean that the force at every point along the wire is zero, but rather that the vector sum of all forces exerted on all parts of the closed wire adds up to zero.

In the study of magnetostatics, all currents flow through closed paths. To understand why, consider the curved wire in Fig. 5-3(b) carrying a current I from point a to point b. In doing so, negative charges accumulate at a, and positive ones at b. The time-varying nature of these charges violates the static assumptions underlying Eqs. (5.1a,b).

If we are interested in the magnetic force exerted on a wire segment l (Fig. 5-3(b)) residing in a uniform magnetic field (while realizing that it is part of a closed current loop), we can integrate Eq. (5.9) to obtain

F m=I(∫ℓ d l)×B=I ℓ×B,(5.12)

where ℓ is the vector directed from a to b (Fig. 5-3(b)). The integral of d l from a to b has the same value irrespective of the path taken between a and b. For a closed loop, points a and b become the same point, in which case ℓ=0 and F m=0.

# Example 5-1: Force on a Semicircular Conductor

The semicircular conductor shown in Fig. 5-4 lies in the x−y plane and carries a current I. The closed circuit is exposed to a uniform magnetic field B=y^B 0. Determine (a) the magnetic force F 1 on the straight section of the wire and (b) the force F 2 on the curved section.

Solution: (a) To evaluate F 1, consider that the straight section of the circuit is of length 2 r and its current flows along the +x direction. Application of Eq. (5.12) with ℓ=x^2 r gives

F 1=x^(2 I r)×y^B 0=z^2 I r B 0 (N).

The z^ direction in Fig. 5-4 is out of the page.

# Figure 5-4 Semicircular conductor in a uniform field (Example 5-1).

Figure 5 4 Full Alternative Text

(b) To evaluate F 2, consider a segment of differential length d l on the curved part of the circle. The direction of d l is chosen to coincide with the direction of the current. Since d l and B are both in the x−y plane, their cross product d l×B points in the negative z direction, and the magnitude of d l×B is proportional to sin ϕ, where ϕ is the angle between d l and B. Moreover, the magnitude of d l is d l=r d ϕ. Hence,

F 2=I∫ϕ=0 π d l×B=−z^I∫ϕ=0 π r B 0 sin ϕ d ϕ=−z^2 I r B 0 (N).

The −z^ direction of the force acting on the curved part of the conductor is into the page. We note that F 2=−F 1, implying that no net force acts on the closed loop, although opposing forces act on its two sections.

Concept Question 5-1: What are the major differences between the electric force F e and the magnetic force F m?

Concept Question 5-2: The ends of a 10-cm long wire carrying a constant current I are anchored at two points on the x axis, namely x=0 and x=6 cm. If the wire lies in the x−y plane in a magnetic field B=y^B 0, which of the following arrangements produces a greater magnetic force on the wire: (a) wire is V-shaped with corners at (0,0), (3,4), and (6,0) or (b) wire is an open rectangle with corners at (0,0), (0,2), (6,2), and (6,0)?

# 5.1.2 Magnetic Torque on a Current-Carrying Loop

When a force is applied on a rigid body that can pivot about a fixed axis, the body will, in general, react by rotating about that axis. The angular acceleration depends on the cross product of the applied force vector F and the distance vector d, which is measured from a point on the rotation axis (such that d is perpendicular to the axis) to the point of application of F (Fig. 5-5). The length of d is called the moment arm, and the cross product

T=d×F (N⋅m)(5.13)

is called the torque. The unit for T is the same as that for work or energy, even though torque does not represent either. The force F applied on the disk shown in Fig. 5-5 lies in the x−y plane and makes an angle θ with d. Hence,

T=z^r F sin θ,(5.14)

where |d|=r, which is the radius of the disk, and F=|F|. From Eq. (5.14), we observe that a torque along the positive z direction corresponds to a tendency for the cylinder to rotate counterclockwise and, conversely, a torque along the −z direction corresponds to clockwise rotation.

# Figure 5-5 The force F acting on a circular disk that can pivot along the z axis generates a torque T=d×F that causes the disk to rotate.

Figure 5 5 Full Alternative Text

We now consider the magnetic torque exerted on a conducting loop under the influence of magnetic forces. We begin with the simple case where the magnetic field B is in the plane of the loop, and then we extend the analysis to the more general case where B makes an angle θ with the surface normal of the loop.

# Magnetic Field in the Plane of the Loop

The rectangular conducting loop shown in Fig. 5-6(a) is constructed from rigid wire and carries a current I. The loop lies in the x−y plane and is allowed to pivot about the axis shown. Under the influence of an externally generated uniform magnetic field B=x^B 0, arms 1 and 3 of the loop are subjected to forces F 1 and F 3 given by

F 1=I(−y^b)×(x^B 0)=z^I b B 0,(5.15a)

and

F 3=I(y^b)×(x^B 0)=−z^I b B 0.(5.15b)

These results are based on the application of Eq. (5.12). We note that the magnetic forces acting on arms 1 and 3 are in opposite directions, and no magnetic force is exerted on either arm 2 or 4 because B is parallel to the direction of the current flowing in those arms.

A bottom view of the loop, depicted in Fig. 5-6(b), reveals that forces F 1 and F 3 produce a torque about the origin O, causing the loop to rotate in a clockwise direction. The moment arm is a/2 for both forces, but d 1 and d 3 are in opposite directions, resulting in a total magnetic torque of

T=d 1×F 1+d 3×F 3=(−x^a 2)×(z^I b B 0)+(x^a 2)×(−z^I b B 0)=y^I a b B 0=y^I A B 0,(5.16)

where A=a b is the area of the loop. The right-hand rule tells us that the sense of rotation is clockwise. The result given by Eq. (5.16) is valid only when the magnetic field B is parallel to the plane of the loop. As soon as the loop starts to rotate, the torque T decreases, and at the end of one quarter of a complete rotation, the torque becomes zero, as discussed next.

# Figure 5-6 Rectangular loop pivoted along the y axis: (a) front view and (b) bottom view. The combination of forces F 1 and F 3 on the loop generates a torque that tends to rotate the loop in a clockwise direction as shown in (b).

Figure 5 6-a Full Alternative Text
Figure 5 6-b Full Alternative Text

# Magnetic Field Perpendicular to the Axis of a Rectangular Loop

For the situation represented by Fig. 5-7, where B=x^B 0, the field is still perpendicular to the loop’s axis of rotation, but because its direction may be at any angle θ with respect to the loop’s surface normal n^, we may now have nonzero forces on all four arms of the rectangular loop. However, forces F 2 and F 4 are equal in magnitude and opposite in direction and are along the rotation axis; hence, the net torque contributed by their combination is zero. The directions of the currents in arms 1 and 3 are always perpendicular to B regardless of the magnitude of θ. Hence, F 1 and F 3 have the same expressions given previously by Eqs. (5.15a,b), and for 0≤θ≤π/2, their moment arms are of magnitude (a/2)sin θ, as illustrated in Fig. 5-7(b). Consequently, the magnitude of the net torque exerted by the magnetic field about the axis of rotation is the same as that given by Eq. (5.16), but modified by sin θ:

T=I A B 0 sin θ.(5.17)

# Figure 5-7 Rectangular loop in a uniform magnetic field with flux density B whose direction is perpendicular to the rotation axis of the loop but makes an angle θ with the loop’s surface normal n^.

Figure 5 7-a Full Alternative Text
Figure 5 7-b Full Alternative Text

According to Eq. (5.17), the torque is maximum when the magnetic field is parallel to the plane of the loop (θ=90°) and zero when the field is perpendicular to the plane of the loop (θ=0). If the loop consists of N turns with each contributing a torque given by Eq. (5.17), then the total torque is

T=N I A B 0 sin θ.(5.18)

The quantity N I A is called the magnetic moment m of the loop. Now, consider the vector

where n^ is the surface normal of the loop and governed by the following right-hand rule: When the four fingers of the right hand advance in the direction of the current I, the direction of the thumb specifies the direction of n^. In terms of m, the torque vector T can be written as

Even though the derivation leading to Eq. (5.20) was obtained for B perpendicular to the axis of rotation of a rectangular loop, the expression is valid for any orientation of B and for a loop of any shape.

Concept Question 5-3: How is the direction of the magnetic moment of a loop defined?

Concept Question 5-4: If one of two wires of equal length is formed into a closed square loop and the other into a closed circular loop, and if both wires are carrying equal currents and both loops have their planes parallel to a uniform magnetic field, which loop would experience the greater torque?