NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 9
Qstn# iv-36 Prvs-QstnNext-Qstn
  • #36
    Figure shows a long U-shaped wire of width l placed in a perpendicular magnetic field B. A wire of length l is slid on the U-shaped wire with a constant velocity v towards right. The resistance of all the wires is r per unit length. At t = 0, the sliding wire is close to the left edge of the U-shaped wire. Draw an equivalent circuit diagram, showing the induced emf as a battery. Calculate the current in the circuit.
    Figure
    Ans :
    The induced emf is given by
    e = Bvl
    Total resistance, R = r × Total length of the wire
    Because the length of the movable wire is l and the distance travelled by the movable wire in time t is vt, the total length of the loop is 2 (l + vt).
    ∴ e = i × 2r (l + vt)
    Bvl = 2ri (l + vt)
    `` \Rightarrow i=\frac{Bvl}{2r(l+vt)}``
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