Irodov 3.289: Two End Resistors Share One Source – Amit & Deepali Sinha Foundation

Problem Irodov 3.289

A rectangular wire frame $ABCD$ has its two end sides $AB$ and $CD$ of resistances $R_1$ and $R_2$ , and its two other sides are resistance-free rails. A straight connector of length $l$ and resistance $R$ lies across the rails, parallel to the end sides, and slides along them with a constant velocity $v$ . The frame lies in a uniform magnetic field of induction $B$ perpendicular to its plane. Neglecting the self-inductance of the circuit, find the current in the connector.

The frame ABCD with resistance R_1 in side AB and R_2 in side CD, and the connector of resistance R sliding along the rails at v. The field (not drawn) is normal to the frame. — The frame $ABCD$ with resistance $R_1$ in side $AB$ and $R_2$ in side $CD$ , and the connector of resistance $R$ sliding along the rails at $\vect v$ . The field (not drawn) is normal to the frame.

Given

$l$
$B$ normal to the frame
$R$ (connector)
$R_1$ ( $AB$ )
$R_2$ ( $CD$ )
$v$ constant
rails of negligible resistance

Find

I

in the connector

Solution

Step 1The connector is the source

Only the connector moves. Its velocity $\vect v$ , its length $\vect l$ and the field $\vect B$ are mutually perpendicular, so the magnetic force on its carriers drives a motional emf along it:

\mathcal{E}_i = (\vect v\times\vect B)\cdot\vect l = Blv \tag{1}

The rest of the frame is at rest and has no emf of its own; the connector acts as a cell of emf (1) with internal resistance $R$ .

The equivalent circuit: the connector as a source E_i with internal resistance R, its ends joined by the ideal rails to R_1 and to R_2, which are in parallel. The source's polarity depends on the field's sense, not given (drawn one way). — The equivalent circuit: the connector as a source $\mathcal{E}_i$ with internal resistance $R$ , its ends joined by the ideal rails to $R_1$ and to $R_2$ , which are in parallel. The source’s polarity depends on the field’s sense, not given (drawn one way).

Step 2The two sides are in parallel

The rails have no resistance, so each rail is a single node: the upper end of the connector, $A$ and $D$ are at one potential, and its lower end, $B$ and $C$ at another. Sides $AB$ and $CD$ therefore both connect the same two nodes, and they are in parallel:

\frac{1}{R_\parallel} = \frac{1}{R_1} + \frac{1}{R_2}, \qquad R_\parallel = \frac{R_1R_2}{R_1 + R_2} \tag{2}

Step 3Ohm’s law for the whole circuit

The emf (1) drives the current through its internal resistance $R$ in series with the parallel pair (2):

I = \frac{\mathcal{E}_i}{R + R_\parallel} = \frac{Blv}{R + \dfrac{R_1R_2}{R_1 + R_2}}

Clearing the compound fraction:

I = \frac{Blv\,(R_1 + R_2)}{R(R_1 + R_2) + R_1R_2}

(3)

Its direction along the connector is that of $\vect v\times\vect B$ ; reversing the field reverses the current but leaves its magnitude unchanged.

Check

Limits. With side $CD$ broken, $R_2 \to \infty$ , (3) becomes $Blv/(R + R_1)$ , the single loop through $R_1$ ; with $AB$ shorted, $R_1 = 0$ , it becomes $Blv/R$ , the connector short-circuited by the rails. Kirchhoff’s loop law round the connector and $AB$ agrees. The current through $AB$ is the share $I_1 = IR_2/(R_1 + R_2)$ , and the voltage drops round that loop add up to the emf:

IR + I_1R_1 = I\left(R + \frac{R_1R_2}{R_1 + R_2}\right) = Blv

Remark

The answer does not depend on where the connector is: with ideal rails the circuit is the same at every position. The flux picture says the same thing more subtly. The flux through the left part of the frame grows at the rate $Blv$ while that through the right part falls at the same rate; each loop has the full emf $Blv$ , and the two loop currents add in the connector.