Two parallel rails, 1 and 2, having negligible resistance are l=10cm apart. They are connected by a 5Ω resistor. The two metal rods ab and cd can slide on the rails. The rods have resistance 10Ω and 15Ω respectively. The rods are moved with constant velocities of 4m/s and 2m/s parallel to the rails, as shown. A uniform magnetic field B = 10 mT exists in the entire space perpendicular to the figure. The current x mA in the 5Ω resistor then calculate 55 x.

Question

Two parallel rails, 1 and 2, having negligible resistance are l=10cm apart. They are connected by a 5Ω resistor. The two metal rods ab and cd can slide on the rails. The rods have resistance 10Ω and 15Ω respectively. The rods are moved with constant velocities of 4m/s and 2m/s parallel to the rails, as shown. A uniform magnetic field B = 10 mT exists in the entire space perpendicular to the figure. The current x mA in the 5Ω resistor then calculate 55 x.

Infinity Learn · Accepted Answer

EMF induced in rod ab is E1=Bv1L sin600Here L = length of abLsin600=l=10 cm=separation between the rails∴E1=Bv1l=10×10−3×4×0.1=4 mVThe equivalent circuit is as shown. Current distribution is also shown. We will apply Kirchhoff’s loop law for the two loops. For loop abfe : 5i+10i1=4For loop cdef: -15(i-i1)-2-5i=0Solving (i) and (ii), gives i=855mA[In equations (i) and (ii), the voltages in mV]

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