# 6.2 Stationary Loop in a Time-Varying Magnetic Field

The stationary, single-turn, conducting, circular loop with contour C and surface area S shown in Fig. 6-2(a) is exposed to a time-varying magnetic field B(t). As stated earlier, the emf induced when S is stationary and the field is time varying is called the transformer emf and is denoted V e m f t r. Since the loop is stationary, d/d t in Eq. (6.6) now operates on B(t) only. Hence,

where the full derivative d/d t has been moved inside the integral and changed into a partial derivative ∂/∂t to signify that it operates on B only. The transformer emf is the voltage difference that would appear across the small opening between terminals 1 and 2, even in the absence of the resistor R. That is, V e m f t r=V 12, where V 12 is the open-circuit voltage across the open ends of the loop. Under dc conditions, V e m f t r=0. For the loop shown in Fig. 6-2(a) and the associated definition for V e m f t r given by Eq. (6.8), the direction of d s, which is the loop’s differential surface normal, can be chosen to be either upward or downward. The two choices are associated with opposite designations of the polarities of terminals 1 and 2 in Fig. 6-2(a).

# Figure 6-2 (a) Stationary circular loop in a changing magnetic field B(t) and (b) its equivalent circuit.

(a) Loop in a changing B field

Figure 6 2-a Full Alternative Text

(b) Equivalent circuit

Figure 6 2-b Full Alternative Text

If the loop has an internal resistance R i, the circuit in Fig. 6-2(a) can be represented by the equivalent circuit shown in Fig. 6-2(b), in which case the current I flowing through the circuit is given by

I=V e m f t r R+R i.(6.9)

For good conductors, R i usually is very small, and it may be ignored in comparison with practical values of R.

The current I induces a magnetic field of its own, B i n d, with a corresponding flux Φ i n d. The direction of B i n d is governed by the right-hand rule: If I is in a clockwise direction, then B i n d points downward through S; conversely, if I is in a counterclockwise direction, then B i n d points upward through S. If the original field B(t) is increasing, which means that d Φ/d t>0, then according to Lenz’s law I has to be in the direction shown in Fig. 6-2(a) in order for B i n d to be in opposition to B(t). Consequently, terminal 2 would be at a higher potential than terminal 1, and V e m f t r would have a negative value. However, if B(t) were to remain in the same direction but to decrease in magnitude, then d Φ/d t would become negative, the current would have to reverse direction, and its induced field B i n d would be in the same direction as B(t) in order to oppose the change (decrease) of B(t). In that case, V e m f t r would be positive.

Despite the presence of the small opening between terminals 1 and 2 of the loop in Fig. 6-2(a), we shall treat the loop as a closed path with contour C. We do this in order to establish the link between B and the electric field E associated with the induced emf, V e m f t r. Also, at any point along the loop, the field E is related to the current I flowing through the loop. For contour C, V e m f t r is related to E by

V e m f t r=∮C E⋅d l.(6.10)

For N=1 (a loop with one turn), equating Eqs. (6.8) and (6.10) gives

∮C E⋅d l=−∫S∂B∂t⋅d s,(6.11)

which is the integral form of Faraday’s law given in Table 6-1. We should keep in mind that the direction of the contour C and the direction of d s are related by the right-hand rule.

By applying Stokes’s theorem to the left-hand side of Eq. (6.11), we have

∫S(∇×E)⋅d s=−∫S∂B∂t⋅d s,(6.12)

and in order for the two integrals to be equal for all possible choices of S, their integrands must be equal, which gives

This differential form of Faraday’s law states that a time-varying magnetic field induces an electric field E whose curl is equal to the negative of the time derivative of B. Even though the derivation leading to Faraday’s law started out by considering the field associated with a physical circuit, Eq. (6.13) applies at any point in space, whether or not a physical circuit exists at that point.

# Example 6-1: Inductor in a Changing Magnetic Field

An inductor is formed by winding N turns of a thin conducting wire into a circular loop of radius a. The inductor loop is in the x−y plane with its center at the origin and connected to a resistor R, as shown in Fig. 6-3. In the presence of a magnetic field B=B 0(y ˆ 2+z ˆ 3)sin ω t, where ω is the angular frequency, find

1. the magnetic flux linking a single turn of the inductor,

2. the transformer emf given that N=10, B 0=0.2 T, a=10 cm, and ω=10 3 rad/s,

3. the polarity of V e m f t r at t=0, and

4. the induced current in the circuit for R=1 k Ω (assume the wire resistance to be much smaller than R).

# Figure 6-3 Circular loop with N turns in the x−y plane. The magnetic field is B=B 0(y ˆ 2+z ˆ 3)sin ω t (Example 6-1).

Figure 6 3 Full Alternative Text

Solution: (a) With d s defined upwards (along +z ˆ direction), the magnetic flux linking each turn of the inductor is

Φ=∫S B . d s=∫S[B 0(y^2+z^3)sin ω t] ⋅z^ d s=3 π a 2 B 0 sin ω t.

(b) To find V e m f t r, we can apply Eq. (6.8) or apply the general expression given by Eq. (6.6) directly. The latter approach gives

V e m f t r=−N d Φ d t=−d d t(3 π N a 2 B 0 sin ω t)=−3 π N ω a 2 B 0 cos ω t.

Having chosen the direction of d s to be upwards, the polarity of V e m f t r is governed by the right-hand rule requiring the direction of the contour C in Fig. 6-3 to pass across the gap from the positive terminal of V e m f t r to the negative terminal. That is,

V e m f t r=V 1−V 2=−3 π N ω a 2 B 0 cos ω t.

For N=10, a=0.1 m, ω=10 3 rad/s, and B 0=0.2 T,

V e m f t r=−188.5 cos 10 3 t (V).

(c) The flux passing through each loop in the upward direction is Φ(t)=3 π a 2 B 0 sin ω t. At t=0, Φ(0)=0, but

d Φ d t t=0=3 π a 2 ω B 0 cos ω t 0|t=0=3 π a 2 ω B 0>0.

Hence, even though the flux itself is zero at t=0, its time derivative is positive. Therefore, the flux is increasing at t=0. Lenz’s law requires the direction of the current I to be such that the flux it induces, Φ i n d, opposes the change in Φ. Since Φ is defined to be upwards, and it is increasing at t=0, Φ i n d should be downward through the loop in order to slow down the increase in Φ. Hence, I should be in the direction shown in Fig. 6-3.

Since I flows through a resistor from its positive voltage terminal to its negative voltage terminal, it follows that V 12 should be negative at t=0, which it is:

V e m f t r=V 1−V 2=−188.5 (V).

(d) The current I is given by

I=V 2−V 1 R=188.5 10 3 cos 10 3 t=0.19 cos 10 3 t (A).

# Example 6-2: Lenz’s Law

Determine voltages V 1 and V 2 across the 2 Ω and 4 Ω resistors shown in Fig. 6-4. The loop is located in the x−y plane, its area is 4 m 2, the magnetic flux density is B=−z ˆ 0.3 t (T), and the internal resistance of the wire may be ignored.

# Figure 6-4 Circuit for Example 6-2.

Figure 6 4 Full Alternative Text

Solution: The flux flowing through the loop is

Φ=∫S B⋅d s=∫S(−z ˆ 0.3 t)⋅z ˆ d s=−0.3 t×4=−1.2 t(Wb),

and the corresponding transformer emf is

V e m f t r=−d Φ d t=1.2 (V).

Since the magnetic flux through the loop is along the −z direction (into the page) and increases in magnitude with time t, Lenz’s law states that the induced current I should be in a direction such that the magnetic flux density B i n d it induces counteracts the direction of change of Φ. Hence, I has to be in the direction shown in the circuit because the corresponding B i n d is along the +z direction in the region inside the loop area. This, in turn, means that V 1 and V 2 are positive voltages.

The total voltage of 1.2 V is distributed across two resistors in series. Consequently,

I=V e m f t r R 1+R 2=1.2 2+4=0.2 A,

and

V 1=I R 1=0.2×2=0.4 V,V 2=I R 2=0.2×4=0.8 V.

Concept Question 6-1: Explain Faraday’s law and the function of Lenz’s law.

Concept Question 6-2: Under what circumstances is the net voltage around a closed loop equal to zero?

Concept Question 6-3: Suppose the magnetic flux density linking the loop of Fig. 6-4 (Example 6-2) is given by B=−z ˆ 0.3 e−t (T). What would the direction of the current be, relative to that shown in Fig. 6-4, for t≥0?