# 6.4 Moving Conductor in a Static Magnetic Field

Consider a wire of length l moving across a static magnetic field B=z ˆ B 0 with constant velocity u, as shown in Fig. 6-7. The conducting wire contains free electrons. From Eq. (5.3), the magnetic force F m acting on a particle with charge q moving with velocity u in a magnetic field B is

F m=q(u×B).(6.22)

# Figure 6-7 Conducting wire moving with velocity u in a static magnetic field.

Figure 6 7 Full Alternative Text

This magnetic force is equivalent to the electrical force that would be exerted on the particle by the electric field E m given by

E m=F m q=u×B.(6.23)

The field E m generated by the motion of the charged particle is called a motional electric field and is in the direction perpendicular to the plane containing u and B. For the wire shown in Fig. 6-7, E m is along −y ˆ. The magnetic force acting on the (negatively charged) electrons in the wire causes them to drift in the direction of −E m; that is, toward the wire end labeled 1 in Fig. 6-7. This, in turn, induces a voltage difference between ends 1 and 2, with end 2 being at the higher potential. The induced voltage is called a motional emf, V e m f m, and is defined as the line integral of E m between ends 2 and 1 of the wire,

V e m f m=V 12=∫2 1 E m⋅d l=∫2 1(u×B)⋅d l.(6.24)

For the conducting wire, u×B=x ˆ u×z ˆ B 0=−y ˆ u B 0 and d l=y ˆ d l. Hence,

V e m f m=V 12=−u B 0 l.(6.25)

In general, if any segment of a closed circuit with contour C moves with a velocity u across a static magnetic field B, then the induced motional emf is given by

# Example 6-3: Sliding Bar

The rectangular loop shown in Fig. 6-8 has a constant width l, but its length x 0 increases with time as a conducting bar slides with uniform velocity u in a static magnetic field B=z ˆ B 0 x. Note that B increases linearly with x. The bar starts from x=0 at t=0. Find the motional emf between terminals 1 and 2 and the current I flowing through the resistor R. Assume that the loop resistance R i≪R.

# Figure 6-8 Sliding bar with velocity u in a magnetic field that increases linearly with x; that is, B=z ˆ B 0 x (Example 6-3).

Figure 6 8 Full Alternative Text

Solution: This problem can be solved by using the motional emf expression given by Eq. (6.26) or by applying the general formula of Faraday’s law. We now show that the two approaches yield the same result.

(a) Motional emf

The sliding bar, being the only part of the circuit that crosses the lines of the field B, is the only part of contour 2341 that contributes to V e m f m. Hence, at x=x 0, for example,

V e m f m=V 12=V 43=∫3 4(u×B)⋅d l=∫3 4(x ˆ u×z ˆ B 0 x 0)⋅y ˆ d l=−u B 0 x 0 l.

The length of the loop is related to u by x 0=u t. Hence,

V e m f m=−B 0 u 2 l t (V).(6.27)

(b) Total emf

Since B is static, V e m f t r=0 and V e m f=V e m f m only. To verify that the same result can be obtained by the general form of Faraday’s law, we evaluate the flux Φ through the surface of the loop. Thus,

Φ=∫S B⋅d s=∫S(z ˆ B 0 x)⋅z ˆ d x d y=B 0 l∫0 x 0 x d x=B 0 l x 0 2 2.(6.28)

Substituting x 0=u t in Eq. (6.28) and then evaluating the negative of the derivative of the flux with respect to time gives

V e m f=−d Φ d t=−d d t B 0 l u 2 t 2 2=−B 0 u 2 l t(V),(6.29)

which is identical with Eq. (6.27). Since V 12 is negative, the current I=B 0 u 2 l t/R flows in the direction shown in Fig. 6-8.

# Example 6-4: Moving Loop

The rectangular loop shown in Fig. 6-9 is situated in the x−y plane and moves away from the origin with velocity u=y ˆ 5 (m/s) in a magnetic field given by

B(y)=z ˆ 0.2 e−0.1 y (T).

# Figure 6-9 Moving loop of Example 6-4.

Figure 6 9 Full Alternative Text

If R=5 Ω, find the current I at the instant that the loop sides are at y 1=2 m and y 2=2.5 m. The loop resistance may be ignored.

Solution: Since u×B is along x ˆ, voltages are induced across only the sides oriented along x ˆ, namely the sides linking point 1 to 2 and point 3 to 4. Had B been uniform, the induced voltages would have been the same, and the net voltage across the resistor would have been zero. In the present case, however, B decreases exponentially with y, thereby assuming a different value over side 1–2 than over side 3–4. Side 1–2 is at y 1=2 m, and the corresponding magnetic field is

B(y 1)=z ˆ 0.2 e−0.1 y 1=z ˆ 0.2 e−0.2 (T).

The induced voltage V 12 is then given by

V 12=∫2 1[u×B(y 1)]⋅d l=∫l/2−l/2(y ˆ 5×z ˆ 0.2 e−0.2)⋅x ˆ d x=−e−0.2 l=−2 e−0.2=−1.637 (V).

Similarly,

V 43=−u B(y 2)l=−5×0.2 e−0.25×2=−1.558 (V).

Consequently, the current is in the direction shown in the figure and its magnitude is

I=V 43−V 12 R=0.079 5=15.8 (m A).

# Example 6-5: Moving Rod Next to a Wire

The wire shown in Fig. 6-10 carries a current I=10 A. A 30-cm long metal rod moves with a constant velocity u=z ˆ 5 m/s. Find V 12, where 1 and 2 are the two ends of the rod.

# Figure 6-10 Moving rod of Example 6-5.

Figure 6 10 Full Alternative Text

Solution: The current I induces a magnetic field

B = ϕ^μ 0 I 2 π r,

where r is the radial distance from the wire and the direction of ϕ^ is into the page on the rod side of the wire. The movement of the rod in the presence of the field B induces a motional emf given by

V 12=∫40 c m 10 c m(u×B)⋅d l=∫40 c m 10 c m z ˆ 5×ϕ^μ 0 I 2 π r⋅r ˆ d r=−5 μ 0 I 2 π∫40 c m 10 c m d r r=−5×4 π×10−7×10 2 π×ln 10 40=13.9 (μ V).

Concept Question 6-4: Suppose that no friction is involved in sliding the conducting bar of Fig. 6-8 and that the horizontal arms of the circuit are very long. Hence, if the bar is given an initial push, it should continue moving at a constant velocity, and its movement generates electrical energy in the form of an induced emf, indefinitely. Is this a valid argument? If not, why not? Can we generate electrical energy without having to supply an equal amount of energy by other means?

Concept Question 6-5: Is the current flowing in the rod of Fig. 6-10 a steady current? Examine the force on a charge q at ends 1 and 2 and compare.