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A loop ABCD containing two resistors as shown in figure is placed in a uniform magnetic field B directed outward to the plane of page. A sliding conductor EF of length l and of negligible resistance moves to the right with a uniform velocity v as shown in Fig. Determine the current in each branch. |
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Answer» Solution : The magnetic field induction B, length and the velocity V of the conductor EF are mutually perpendicular, hence the emf induced in it is e = Bly (with end F of the rod at higher potential) . ` therefore ` The effective electric circuit can be redrawn as SHOWN in Fig  The resistances `R_1` and `R_2`are in parallel, so the equivalent resistance R is given by ` 1/R = (1)/(R_1) + (1)/(R_2)` From Ohms law, the total current is `i = e/R` `i = Bvl ((1)/(R_1) + (1)/(R_2))` Current in AD is `i_1 = (BLV)/(R_1) , ` current in BC is `i_2 = (Blv)/(R_2)` |
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