Worked examples · Sheet 1

Electromotive forces induced by varying fields

Theory used: Magnetic flux linkage Faraday-Neumann-Lenz law Induced current Induced charge (Felici's law)Direction of the induced current (Lenz's law)

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A coil of $N = 200$ circular turns of radius $r = 3.0 cm$ has its axis parallel to a uniform magnetic field, whose magnitude increases uniformly from $0$ to $0.50 T$ during the interval $Δ t = 0.10 s$ . The coil is closed on a circuit of total resistance $R = 4.0 Ω$ , and its self-induction is negligible. Compute the magnitude of the induced emf and of the current.

Φ_{B} = N B π r^{2}, E = - \frac{d Φ _{B}}{d t}, i = \frac{∣ E ∣}{R}

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

π r^{2} ∣ E ∣ i = π \cdot 0.03 0^{2} m^{2} \approx 2.83 \cdot 1 0^{- 3} m^{2} = 200 \cdot 2.83 \cdot 1 0^{- 3} \cdot \frac{0.50}{0.10} V \approx 2.8 V = \frac{2.83}{4.0} A \approx 0.71 A

The axis of the coil is parallel to the field, so the normal to the turns makes a zero angle with $B$ and the flux linkage is $N B$ times the area of one turn. Only the magnitude of the field varies, and it varies uniformly, therefore the derivative of the flux coincides with the ratio $\frac{Δ Φ _{B}}{Δ t}$ and the Faraday-Neumann-Lenz law gives the constant emf $N π r^{2} \frac{Δ B}{Δ t}$ . The current is obtained by dividing by the resistance, since the self-induction is negligible.

Dimensional check: $T \cdot m^{2} /s = Wb/s$ , and one weber per second is one volt, in agreement with Faraday’s law. The induced current stops when the field stops increasing.

Exercise 2

A square loop of side $ℓ = 20 cm$ and resistance $R = 0.50 Ω$ is at rest in a uniform magnetic field, with its normal inclined at $θ = 6 0^{\circ}$ to the field. The magnitude of the field varies according to the law $B (t) = α t^{2}$ , with $α = 0.40 T/ s^{2}$ . Neglecting self-induction, compute the induced current at time $t_{1} = 3.0 s$ and the charge that passes through the loop between $t = 0$ and $t_{1}$ .

Φ_{B} (t) = α t^{2} ℓ^{2} cos θ, E (t) = - \frac{d Φ _{B}}{d t} = - 2 α t ℓ^{2} cos θ

∣ i (t_{1}) ∣ = \frac{2 α t _{1} ℓ ^{2} cos θ}{R}, ∣ q ∣ = \frac{∣ Φ _{B} ( t _{1} ) - Φ _{B} ( 0 ) ∣}{R} = \frac{α t _{1}^{2} ℓ ^{2} cos θ}{R}

∣ E (t_{1}) ∣ ∣ i (t_{1}) ∣ ∣ q ∣ = 2 \cdot 0.40 \cdot 3.0 \cdot 0.2 0^{2} \cdot 0.50 V = 0.048 V = \frac{0.048}{0.50} A \approx 0.096 A = \frac{0.40 \cdot 3. 0 ^{2} \cdot 0.2 0 ^{2} \cdot 0.50}{0.50} C \approx 0.14 C

The loop is flat and the field is uniform, so the flux is the product of the field, the area $ℓ^{2}$ and $cos θ$ . The area and the angle are constant, therefore the flux varies only through $B (t)$ , and its derivative is $2 α t ℓ^{2} cos θ$ . The emf grows linearly in time, and at time $t_{1}$ it equals $48 mV$ .

For the charge we apply Felici’s law: the charge that passes through the loop depends only on the change in flux between the initial instant, at which the flux is zero, and the instant $t_{1}$ . The same result is obtained by integrating $∣ i (t) ∣ = \frac{2 α t ℓ ^{2} cos θ}{R}$ between $0$ and $t_{1}$ , since $\int_{0}^{t_{1}} 2 t d t = t_{1}^{2}$ .

Exercise 3

A long solenoid, with $n = 2000$ turns per meter, contains inside it, far from the ends, a small coaxial coil of $N = 100$ turns of radius $r = 1.5 cm$ . The current in the solenoid increases uniformly from $0$ to $3.0 A$ in $50 ms$ . Compute the emf induced in the coil and determine the direction of the induced current relative to that of the current in the solenoid.

B = μ_{0} n i_{s}, Φ_{B} = N B π r^{2} = N μ_{0} n π r^{2} i_{s}, ∣ E ∣ = N μ_{0} n π r^{2} \frac{Δ i _{s}}{Δ t}

μ_{0} n π r^{2} ∣ E ∣ = 4 π \cdot 1 0^{- 7} \cdot 2000 T/A \approx 2.51 \cdot 1 0^{- 3} T/A = π \cdot 0.01 5^{2} m^{2} \approx 7.07 \cdot 1 0^{- 4} m^{2} = 100 \cdot 2.51 \cdot 1 0^{- 3} \cdot 7.07 \cdot 1 0^{- 4} \cdot \frac{3.0}{0.050} V \approx 1.1 \cdot 1 0^{- 2} V

Inside a long solenoid the field is uniform, parallel to the axis and equal to $μ_{0} n i_{s}$ , as seen in the topic on sources of the magnetic field. The coil is coaxial, so its turns are perpendicular to the field and the flux linkage is $N B π r^{2}$ , proportional to the current $i_{s}$ . The current increases uniformly by $60 A$ per second, and the Faraday-Neumann-Lenz law gives a constant emf of about $11 mV$ .

The flux in the coil increases, therefore by Lenz’s law the induced current generates a field opposite to that of the solenoid; hence the current in the coil flows in the direction opposite to that of the solenoid. During the increase the field in the solenoid reaches $μ_{0} n \cdot 3.0 A \approx 7.5 mT$ , a typical value for laboratory solenoids.