Two parallel rails with negligible resistance are 10.0 cm apart. They are connected by a $5 Ω$ resistor. The circuit also contains two metal rods having resistances of $10.0 Ω$ and $15.0 Ω$ along the rails. The rods are pulled away from the resistor at constant speeds $4.00 m / s$ and $2.00 m / s$ respectively. A uniform magnetic field of magnitude $0.01 T$ is applied perpendicular to the, plane of rails. Determine the current in the $5.0 Ω$ resistor.

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$\frac{8}{22} m A$

$\frac{8}{44} m A$

$\frac{8}{33} m A$

$\frac{8}{55} m A$

answer is D.

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Detailed Solution

In the figures $R = 5.0 Ω, r_{1} = 10 Ω, r_{2} = Ω, e_{1} = 4 \times 10^{- 3} V a n d e_{2} = 2 \times 10^{- 3} V$

Refer Figure (b) Net resistance of the circuit $= r_{2} + \frac{R r_{1}}{R + r_{1}} = 15 + \frac{10 \times 15}{10 + 15} = \frac{55}{3} Ω$

current, $i = \frac{e_{2}}{N e t r e s i s \tan c e} = \frac{2 \times 10^{- 3}}{\frac{55}{3}} = \frac{6}{55} \times 10^{- 3} A$

Current through R,

$i_{1} = (\frac{r_{1}}{R + r_{1}}) i = (\frac{10}{10 + 5}) (\frac{6}{55} \times 10^{- 3}) A$

$= \frac{4}{55} \times 10^{- 3} A = \frac{4}{55} m A$

Refer figure (b) Net resistance of the circuit $= r_{1} + \frac{R r_{2}}{R + r_{2}} = 10 + \frac{5 \times 15}{5 + 15} = \frac{55}{4} Ω$

Current,

$i' = \frac{e_{2}}{N e t r e s i s \tan c e} = \frac{4 \times 10^{- 3}}{\frac{55}{4}} = \frac{16}{55} \times 10^{- 3} A$

Current through R,

$i'_{1} = (\frac{r_{2}}{R + r_{2}}) i = (\frac{15}{15 + 5}) (\frac{16}{55} \times 10^{- 3}) A = \frac{12}{55} m A$

From superposition principle net current through $5.0 Ω$ resistor is

$i'_{1} - i_{1} = \frac{8}{55} m A$ from $d$ to $c$ .

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