Two long horizontal rails, a distance l = 0.50 m apart, are joined at one end by a resistor of…

Two long horizontal rails, a distance $l = 0.50\ \text{m}$ apart, are joined at one end by a resistor of resistance $R = 0.80\ \Omega$ . A uniform magnetic field of $B = 0.40\ \text{T}$ is vertical, into the page in the figure. A straight metal rod of mass $m = 0.060\ \text{kg}$ lies across the rails and slides on them without friction. A light string joins the middle of the rod to a block of mass $M = 0.040\ \text{kg}$ , which hangs over a smooth pulley at the edge of the table. The rod is released from rest. The resistance of the rod and of the rails can be ignored, and the string stays tight. Take $g = 9.8\ \text{m/s}^2$ .

Top view of two long horizontal parallel rails, a distance l apart, joined at the left end by a resistor R. A metal rod lies across the rails at right angles, free to slide without friction. A string tied to the middle of the rod runs to the right to a pulley at the edge of the table, and a block of mass M hangs below the table from the other end of the string. Crosses over the whole area between the rails show a uniform magnetic field B going into the page.

What is the acceleration of the rod at the instant it is released?
What terminal speed does the rod reach?
What is the speed of the rod when its acceleration has fallen to one quarter of its initial value?
At what rate is heat produced in the resistor at the terminal speed?
What charge flows through the resistor while the rod moves its first $2.0\ \text{m}$ from rest?

Decide first which forces act on the rod and which on the block, and how the motion of the rod makes a current.

The worked solution

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Given. $l = 0.50\ \text{m}$ , $R = 0.80\ \Omega$ , $B = 0.40\ \text{T}$ , $m = 0.060\ \text{kg}$ , $M = 0.040\ \text{kg}$ , $g = 9.8\ \text{m/s}^2$ .

Plan. When the rod moves at speed $v$ , the emf is $B l v$ and the current is $I = \dfrac{B l v}{R}$ . The magnetic force on the rod is $B I l = \dfrac{B^2 l^2 v}{R}$ , and by Lenz's law it opposes the motion. Put $k = \dfrac{B^2 l^2}{R} = \dfrac{(0.40\ \text{T})^2 (0.50\ \text{m})^2}{0.80\ \Omega} = 0.050\ \text{N s/m}$ . The rod and block move together, so $M g - k v = (m + M)\,a$ .

(a) At release. The speed is zero, so there is no current and no magnetic force. $a_0 = \dfrac{M g}{m + M} = \dfrac{0.040\ \text{kg} \times 9.8\ \text{m/s}^2}{0.100\ \text{kg}} = 3.92\ \text{m/s}^2$ .

(b) Terminal speed. The acceleration is zero when $k v_t = M g$ . So $v_t = \dfrac{M g}{k} = \dfrac{0.392\ \text{N}}{0.050\ \text{N s/m}} = 7.84\ \text{m/s}$ .

(c) A quarter of the first acceleration. The acceleration is $\dfrac{M g - k v}{m + M}$ . It is $\dfrac{a_0}{4}$ when the magnetic force is $\dfrac{3}{4} M g$ , so $v = \dfrac{3}{4} v_t = 5.88\ \text{m/s}$ .

(d) Heating at the terminal speed. $I = \dfrac{B l v_t}{R}$ . With $B l v_t = 0.40 \times 0.50 \times 7.84 = 1.57\ \text{V}$ and $R = 0.80\ \Omega$ , this is $1.96\ \text{A}$ . The rate of heating is $I^2 R = (1.96\ \text{A})^2 \times 0.80\ \Omega = 3.07\ \text{W}$ .

(e) Charge. The flux through the loop of the rod, the rails and the resistor changes by $\Delta\Phi = B\,l\,x = 0.40\ \text{T} \times 0.50\ \text{m} \times 2.0\ \text{m} = 0.40\ \text{Wb}$ . The charge is $\dfrac{\Delta\Phi}{R} = \dfrac{0.40\ \text{Wb}}{0.80\ \Omega} = 0.50\ \text{C}$ , whatever the speed.

Answer: (a) $3.92\ \text{m/s}^2$ . (b) $7.84\ \text{m/s}$ . (c) $5.88\ \text{m/s}$ . (d) $3.07\ \text{W}$ . (e) $0.50\ \text{C}$ .

Check. At the terminal speed the block falls at $7.84\ \text{m/s}$ and loses $M g v_t = 0.392\ \text{N} \times 7.84\ \text{m/s} = 3.07\ \text{W}$ of potential energy per second, all of it turned into heat. This equals $I^2 R$ in (d). The magnetic force $B I l = 0.392\ \text{N}$ equals $M g$ .

Watch out. In (a) the mass is $m + M = 0.100\ \text{kg}$ , not $M$ alone: using $M$ gives $9.8\ \text{m/s}^2$ . In (c) a quarter of the acceleration is not a quarter of the terminal speed: the acceleration falls in step with the speed, so it is $\tfrac{3}{4} v_t$ .

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