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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 8
Qstn# iv-32 Prvs-QstnNext-Qstn
  • #32
    A copper wire bent in the shape of a semicircle of radius r translates in its plane with a constant velocity v. A uniform magnetic field B exists in the direction perpendicular to the plane of the wire. Find the emf induced between the ends of the wire if (a) the velocity is perpendicular to the diameter joining free ends, (b) the velocity is parallel to this diameter.
    Ans : (a) The emf induced between the ends of the wire is calculated using the following formula:
    e = Bv (Effective length of the wire)
    Effective length of the wire = Component of length perpendicular to the velocity
    Here, the component of length moving perpendicular to v is 2r.
    ∴ Induced emf, e = Bv2r (b) When the velocity is parallel to the diameter of the semicircular wire, the component of its length perpendicular to its velocity is zero.
    ∴ Induced emf, e = 0
    Page No 308:
  • #32-a
    the velocity is perpendicular to the diameter joining free ends,
    Ans : The emf induced between the ends of the wire is calculated using the following formula:
    e = Bv (Effective length of the wire)
    Effective length of the wire = Component of length perpendicular to the velocity
    Here, the component of length moving perpendicular to v is 2r.
    ∴ Induced emf, e = Bv2r
  • #32-b
    the velocity is parallel to this diameter.
    Ans : When the velocity is parallel to the diameter of the semicircular wire, the component of its length perpendicular to its velocity is zero.
    ∴ Induced emf, e = 0
    Page No 308: