Worked examples · Sheet 3

Directions of induced currents

Theory used: Direction of the induced current (Lenz's law)lenz variazione Faraday-Neumann-Lenz law Induced current Emf of a moving rod

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A circular loop of radius $r = 5.0 cm$ and resistance $R = 0.10 Ω$ lies in the plane of the page, in a uniform magnetic field pointing out of the page. The magnitude of the field decreases uniformly from $0.60 T$ to $0.20 T$ in $Δ t = 0.20 s$ . Neglecting self-induction, determine the direction of the induced current, as seen by the reader, and its magnitude.

The outward field

B

decreases: the induced current flows counterclockwise and generates inside the outward field

B^{'}

, in the same direction as

B

Φ_{B} = B π r^{2}, ∣ E ∣ = π r^{2} \frac{∣Δ B ∣}{Δ t}, i = \frac{∣ E ∣}{R}

∣ E ∣ i = π \cdot 0.05 0^{2} \cdot \frac{0.60 - 0.20}{0.20} V \approx 1.6 \cdot 1 0^{- 2} V = \frac{1.57 \cdot 1 0 ^{- 2}}{0.10} A \approx 0.16 A

We apply Lenz’s law in three steps. The outward flux through the loop decreases, because the field decreases; the induced field must oppose this decrease, so inside the loop it points outward, in the same direction as the external field. By the right-hand rule, with the thumb pointing toward the reader, the fingers curl counterclockwise; therefore the induced current flows counterclockwise.

The magnitude follows from the Faraday-Neumann-Lenz law. The field changes by $0.40 T$ in $0.20 s$ , that is, at $2.0 T/s$ , and the emf is the product of this rate and the area of the loop. If the field increased at the same rate, the magnitude would be the same and the direction would be clockwise.

Exercise 2

A square loop of side $a = 10 cm$ and resistance $R = 0.020 Ω$ moves with constant velocity $v = 0.50 m/s$ in the plane of the page and crosses a region of width $d = 30 cm$ in which there is a uniform field $B = 0.40 T$ pointing into the page; outside the region the field is zero. The velocity is perpendicular to the edges of the region and at time $t = 0$ the leading side of the loop enters the field. Neglecting self-induction, describe the induced current as a function of time and compute the energy dissipated during the crossing.

E = B a v, i = \frac{B a v}{R}, t_{entry} = \frac{a}{v}, t_{exit start} = \frac{d}{v}, t_{end} = \frac{d + a}{v}

W_{diss} = 2 \cdot R i^{2} \cdot \frac{a}{v} = \frac{2 B ^{2} a ^{3} v}{R}

E i t_{entry} W_{diss} = 0.40 \cdot 0.10 \cdot 0.50 V = 0.020 V = \frac{0.020}{0.020} A = 1.0 A = 0.20 s, t_{exit start} = 0.60 s, t_{end} = 0.80 s = 2 \cdot 0.020 \cdot 1. 0^{2} \cdot 0.20 J = 8.0 \cdot 1 0^{- 3} J

Induced current in amperes, positive counterclockwise, with time in tenths of a second: the current flows only while the loop enters and while it leaves the field

During the entry, which lasts until time $t_{entry}$ , the leading side of the loop is in the field and the trailing side is not; the immersed area grows at the rate $a v$ , so the inward flux grows at the rate $B a v$ , and this is the emf. By Lenz’s law the current generates an outward field, therefore it flows counterclockwise. When the loop is entirely immersed, until time $t_{exit start}$ at which the leading side reaches the opposite edge of the region, the flux is constant, so the current is zero, even though the loop moves in the field. During the exit, which ends at time $t_{end}$ , the inward flux decreases at the same rate, and the current has the same magnitude and clockwise direction.

The current flows during two intervals of $0.20 s$ each, and in each interval the power dissipated is $R i^{2} = 0.020 W$ . The energy dissipated, $8.0 mJ$ , equals the work of the external force that keeps the velocity constant: in the two phases this force balances the magnetic force $B ia = 0.040 N$ on the immersed side, and its work is $0.040 \cdot 0.20 J$ in total, since the loop travels $0.10 m$ in each phase.

Exercise 3

An infinite straight wire carries the current $I = 10 A$ upward. In the plane of the wire, to its right, there is a rectangular loop with the sides of length $b = 20 cm$ parallel to the wire and the sides of length $a = 10 cm$ perpendicular to it. The loop moves away from the wire with velocity $v = 2.0 m/s$ , and at the instant considered the nearer side is at distance $x = 5.0 cm$ from the wire. Compute the induced emf at that instant and determine the direction of the current.

Rectangular loop moving away from a wire carrying the current

I

: the field of the wire points into the page in the region of the loop and the flux decreases, so the induced current

i

flows clockwise

B (s) = \frac{μ _{0} I}{2 π s}, E = (B (x) - B (x + a)) b v = \frac{μ _{0} I ab v}{2 π x ( x + a )}

B (x) B (x + a) E = \frac{4 π \cdot 1 0 ^{- 7} \cdot 10}{2 π \cdot 0.050} T = 4.0 \cdot 1 0^{- 5} T = \frac{4 π \cdot 1 0 ^{- 7} \cdot 10}{2 π \cdot 0.15} T \approx 1.33 \cdot 1 0^{- 5} T = (4.0 - 1.33) \cdot 1 0^{- 5} \cdot 0.20 \cdot 2.0 V \approx 1.1 \cdot 1 0^{- 5} V

The field of the wire is perpendicular to the plane of the page and has magnitude $\frac{μ _{0} I}{2 π s}$ at distance $s$ , as seen in the topic on sources of the magnetic field; with the current upward, to the right of the wire it points into the page. The field is not uniform, so it is convenient to use the motional emf side by side. The two sides perpendicular to the wire move parallel to themselves, and the Lorentz force on their charges is perpendicular to the side, therefore they do not contribute. On the two sides parallel to the wire the emf equals $B b v$ , with the field computed at the distance of each side; in the circuit the two emfs are opposite, so the total emf is their difference.

The direction follows from Lenz’s law: moving away from the wire the loop goes where the field is weaker, therefore the inward flux decreases and the induced current generates an inward field, that is, it flows clockwise. The same result is obtained by differentiating the flux $Φ_{B} = \frac{μ _{0} I b}{2 π} ln \frac{x + a}{x}$ with respect to time, with $\frac{d x}{d t} = v$ .