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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 8

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  • #27-b
    For what electric field inside the rod, the electric force on a free elctron will balance the magnetic force? How is this electric field created?
    Ans : The electrostatic force on the charged particle is qE.
    Here,
    qE = qvB
    ⇒ E = (1 × 10-1 ) × (1 × 10-1)
    = 1 × 10-2 V/m
    It is created because of the induced emf.
  • #27-c
    Find the motional emf between the ends of the rod.
    Ans : Motional emf between the ends of the rod, e = Bvl
    ⇒ e = 0.1 × 0.1 × 0.2
    = 2 × 10-3 V
    Page No 308:
  • Qstn #28
    A metallic metre stick moves with a velocity of 2 m s-1 in a direction perpendicular to its length and perpendicular to a uniform magnetic field of magnitude 0.2 T. Find the emf induced between the ends of the stick.
    Ans : Given:
    Length of the stick, l = 1 m
    Magnetic field, B = 0.2 T
    Velocity of the stick, v = 2 m/s
    Thus, we get
    Induced emf, e = Blv = 0.2 × 1 × 2 = 0.4 V
    Page No 308:
  • Qstn #29
    A 10 m wide spacecraft moves through the interstellar space at a speed 3 × 107 m s-1. A magnetic field B = 3 × 10-10 T exists in the space in a direction perpendicular to the plane of motion. Treating the spacecraft as a conductor, calculate the emf induced across its width.
    Ans : Given:
    l = 10 m
    v = 3 × 107 m/s
    B = 3 × 10-10 T
    Now,
    Motional emf = Bvl
    = (3 × 10-10 ) × (3 × 107 ) × (10)
    = 9 × 10-2
    = 0.09 V
    Page No 308:
  • Qstn #30
    The two rails of a railway track, insulated from each other and from the ground, are connected to a millivoltmeter. What will be the reading of the millivoltmeter when a train travels on the track at a speed of 180 km h-1? The vertical component of earth’s magnetic field is 0.2 × 10-4 T and the rails are separated by 1 m.
    Ans : Here,
    Velocity of the train, v = 180 km/h = 50 m/s
    Earth's magnetic field, B = 0.2 × 10-4 T
    Separation between the railings, l = 1 m
    Induced emf, e = Bvl = 0.2 × 10-4 × 50
    = 10-3 V
    So, the voltmeter will record 1 mV as the reading.
    Page No 308:
  • Qstn #31
    A right-angled triangle abc, made from a metallic wire, moves at a uniform speed v in its plane as shown in figure. A uniform magnetic field B exists in the perpendicular direction. Find the emf induced
  • #31-a
    in the loop abc,
    Ans : The emf induced in loop abc is zero, as there is no change in the magnetic flux through it.
  • #31-b
    in the segment bc,
    Ans : The emf induced is given by
    `` e=\left(\stackrel{\to }{v}\times \stackrel{\to }{B}\right).\stackrel{\to }{l}``
    Emf induced in segment bc, e = Bvl (With positive polarity at point C)
  • #31-c
    in the segment ac and
    Ans : There is no emf induced in segment bc, as the velocity is parallel to its length.
  • #31-d
    in the segment ab.
    Figure
    Ans : The emf induced in segment ab is calculated by the following formula:
    e = B.v. (Effective length of ab)
    The effective length of ab is along the direction perpendicular to its velocity.
    Emf induced, e = B.v.(bc)
    Page No 308:
  • 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
  • #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:
  • Qstn #33
    A wire of length 10 cm translates in a direction making an angle of 60° with its length. The plane of motion is perpendicular to a uniform magnetic field of 1.0 T that exists in the space. Find the emf induced between the ends of the rod if the speed of translation is 20 cm s-1.
    Ans : Given:
    Length of the rod, l = 10 cm = 0.1 m
    Angle between the velocity and length of the rod, θ = 60°
    Magnetic field, B = 1 T
    Velocity of the rod, v = 20 cm/s = 0.2 m/s
    The motional emf induced in the rod is given by
    `` e=\left(\stackrel{\to }{v}\times \stackrel{\to }{B}\right).\stackrel{\to }{l}``
    ∴ e = Bvl sin 60°
    We take the component of the length vector that is perpendicular to the velocity vector.
    ∴ e = `` 1\times 0.2\times 0.1\times \sqrt{\frac{3}{2}}``
    = 17.32 × 10-3 V
    Page No 308:
  • Qstn #34
    A circular copper-ring of radius r translates in its plane with a constant velocity v. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the ring. Consider different pairs of diametrically opposite points on the ring.
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