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Two parallel rails with negligible resistance are 10.0cm apart. They are connected by a 5.0Omegaresistor. The circuit also contains two metal rods having resistances of 10.0Omegaand 15.0Omegaalong 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 the rails. Determine the current in the 5.0Omegaresistor. |
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Answer» Solution :In the figure ` R = 5.0 Omega , r_1 = 10 Omega, r_2 = 15Omega` ` e_1 = Blv_1 = 0.01 xx 0.1 xx 4 = 4 xx 10^(-3) V` ` e_2 = Blv_2 = 0.01 xx 0.1 xx 2 = 2 xx 10^(-3) V` APPLYING kirchoff.s law to the left loop: `10i + 5(i_1 - i_2) = 4.10^(-3)` `rArr 15 i_1 + (5(i_1- i_2) = 4.10^(-3)` ` rArr 15i_1 - 5i_2 = 4 xx 10^(-3) to (1) ` Right loop `: 15i_2 - 5(i_1 - i_2) = 2.10^(-3)` `rArr 20i_2 - 5i_1 = 2 xx 10^(-3) to (2)` SOLVING (1) and (2) gives `i_1 =18/55 xx 10^(-3) A` and `i_2 = 10/55 xx 10^(-3)A` `rArr` current through `5Omega = i_1 - i_2` ` = 8/55 xx 10^(-3) A = 8/55 mA` |
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