Worked examples · Sheet 2

Rods moving on rails

Theory used: Emf of a moving rod Induced current Rod launched on rails Energy balance Direction of the induced current (Lenz's law)Induced charge (Felici's law)

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Two horizontal conducting rails, a distance $ℓ = 0.50 m$ apart, are closed by a resistor $R = 2.0 Ω$ and immersed in a uniform magnetic field $B = 0.80 T$ , perpendicular to the plane of the rails and pointing into the page. A conducting rod, perpendicular to the rails, is pulled to the right by an external force at constant velocity $v = 4.0 m/s$ . Neglecting friction, the resistances of the rails and the rod, and self-induction, compute the induced emf, the current and its direction, the external force, and the power dissipated in the resistor.

The rod pulled to the right: the induced current

I

flows counterclockwise, the magnetic force

F_{m}

points to the left and the external force

F_{ext}

balances it

E = B ℓ v, I = \frac{E}{R}, F_{ext} = F_{m} = I ℓ B = \frac{B ^{2} ℓ ^{2} v}{R}, P = R I^{2}

E I F_{ext} P = 0.80 \cdot 0.50 \cdot 4.0 V = 1.6 V = \frac{1.6}{2.0} A = 0.80 A = 0.80 \cdot 0.50 \cdot 0.80 N = 0.32 N = 2.0 \cdot 0.8 0^{2} W \approx 1.3 W

The rod, the field and the velocity are mutually perpendicular, so the theorem on the emf of a moving rod applies. The circuit has resistance $R$ , therefore the current is $\frac{E}{R}$ . The area of the circuit increases and with it the inward flux; by Lenz’s law the current generates an outward field inside the circuit, and therefore flows counterclockwise, that is, from bottom to top in the rod.

The current-carrying rod is subject to the magnetic force $I ℓ \times B$ , pointing to the left. The velocity is constant, so the net force is zero and the external force has the same magnitude as the magnetic force. As a check, the power of the external force is $F_{ext} v = 0.32 \cdot 4.0 W = 1.28 W$ , equal to the power dissipated in the resistor, in agreement with the energy balance.

Exercise 2

Two vertical conducting rails, a distance $ℓ = 25 cm$ apart, are connected at the top by a resistor $R = 0.50 Ω$ and immersed in a uniform horizontal magnetic field $B = 0.60 T$ , perpendicular to the plane of the rails. A horizontal rod of mass $m = 20 g$ is released from rest, remaining in frictionless contact with the rails. Neglecting the resistances of the rails and the rod and self-induction, determine the law $v (t)$ , the terminal velocity, and the time after which the velocity has reached $90 %$ of the terminal velocity.

m \frac{d v}{d t} = m g - \frac{B ^{2} ℓ ^{2}}{R} v ⟹ \frac{d v}{d t} + \frac{1}{τ} v = g, τ = \frac{m R}{B ^{2} ℓ ^{2}}

v (t) = v_{lim} (1 - e^{- \frac{t}{τ}}), v_{lim} = g τ = \frac{m g R}{B ^{2} ℓ ^{2}}, t_{90} = τ ln 10

B^{2} ℓ^{2} τ v_{lim} t_{90} = 0.6 0^{2} \cdot 0.2 5^{2} T^{2} \cdot m^{2} = 0.0225 T^{2} \cdot m^{2} = \frac{0.020 \cdot 0.50}{0.0225} s \approx 0.44 s = \frac{0.020 \cdot 9.81 \cdot 0.50}{0.0225} m/s \approx 4.4 m/s = 0.444 \cdot 2.303 s \approx 1.0 s

The rod is subject to its weight, downward, and to the magnetic force due to the induced current, of magnitude $\frac{B ^{2} ℓ ^{2} v}{R}$ as in the theorem on the rod on rails; by Lenz’s law the magnetic force opposes the fall, so it points upward. Newton’s second law gives a first-order linear equation with $a = \frac{1}{τ}$ and $b = g$ , and with the condition $v (0) = 0$ the solution tends to $\frac{b}{a} = g τ$ .

The terminal velocity is the one at which the magnetic force balances the weight. The condition $v = 0.90 v_{lim}$ is equivalent to $e^{- \frac{t}{τ}} = 0.10$ , that is, to $t = τ ln 10$ ; after about one second the rod therefore falls at practically constant velocity. In steady state the power of the weight, $m g v_{lim} \approx 0.86 W$ , is entirely dissipated in the resistor.

Exercise 3

On two horizontal rails, a distance $ℓ = 0.50 m$ apart and closed by a resistor $R = 0.20 Ω$ , a rod of mass $m = 0.10 kg$ is launched with velocity $v_{0} = 3.0 m/s$ in a uniform field $B = 0.40 T$ perpendicular to the plane of the rails. Neglecting friction, the resistances of the rails and the rod, and self-induction, compute the time constant of the motion, the distance covered until the rod stops, the heat dissipated in the resistor, and the charge that passes through it.

τ = \frac{m R}{B ^{2} ℓ ^{2}}, x_{tot} = \int_{0}^{+ \infty} v_{0} e^{- \frac{t}{τ}} d t = v_{0} τ, W_{R} = \frac{1}{2} m v_{0}^{2}, q = \frac{B ℓ x _{tot}}{R}

τ x_{tot} W_{R} q = \frac{0.10 \cdot 0.20}{0.4 0 ^{2} \cdot 0.5 0 ^{2}} s = 0.50 s = 3.0 \cdot 0.50 m = 1.5 m = \frac{1}{2} \cdot 0.10 \cdot 3. 0^{2} J = 0.45 J = \frac{0.40 \cdot 0.50 \cdot 1.5}{0.20} C = 1.5 C

The only force acting along the rails is the magnetic braking force, so by the theorem on the rod on rails the velocity decreases exponentially with time constant $τ$ . The distance covered is the integral of the velocity over the whole motion, and equals $v_{0} τ$ .

By the energy balance the power dissipated in the resistor equals at every instant the power lost by the rod, so the total heat is the initial kinetic energy. The charge follows from Felici’s law: the flux linkage changes by $B ℓ x_{tot}$ , because the area of the circuit increases by $ℓ x_{tot}$ . As a check, the impulse-momentum theorem gives the same value, since the impulse of the magnetic force is $\int B ℓ i d t = B ℓ q$ and must cancel the initial momentum, therefore $q = \frac{m v _{0}}{B ℓ} = \frac{0.30}{0.20} C = 1.5 C$ .