Two parallel rails with negligible resistance are 15.0 cm and are connected by a

Question

Two parallel rails with negligible resistance are 15.0 cm and are connected by a resistor of resistance R_3 = 5.00 Ohm. The circuit also contains two metal rods having resistances of R_1 = 11.5 Ohm and R_2 = 15.0 Ohm sliding along the rails (see figure). The rods are pulled away from the resistor at constant speeds of v_1 = 4.00 m/s and v_2 = 2.00 m/s, respectively. A uniform magnetic field of magnitude B = 0.0100 T is applied perpendicular to the plane of the rails. Determine the current in R_3.

Explanation / Answer

induced emf in the first rod, V1 = B*v1*L

= 0.01*4*0.15

= 0.006 volts

current through R1, I1 = V1/(R1+R3)

= 0.006/(11.5 + 5)

= 3.63*10^-4 A

induced emf in the second rod, V2 = B*v2*L

= 0.01*2*0.15

= 0.003 volts

current through R2, I2 = V2/(R2+R3)

= 0.003/(15 + 5)

= 1.5*10^-4 A

so, current through R3 = I1 + I2

= 3.63*10^-4 + 1.5*10^-4

= 5.13*10^-4 0.513 mA or 513 micro A<<<<<<-----------------Answer

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