Electromagnetic Induction


Electromagnetic Induction

  1. What is the self-inductance of a coil which produces 5V when the current changes from 3 ampere to 2 ampere in one millisecond?









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    L =
    e
    di
    dt

    =
    e dt
    =
    5 × 10-3
    H = 5 mH
    di(3 - 2)

    Correct Option: B

    L =
    e
    di
    dt

    =
    e dt
    =
    5 × 10-3
    H = 5 mH
    di(3 - 2)


  1. ​If N is the number of turns in a coil, the value of self inductance varies as​​









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    L =
    ; φ = BA;
    i

    B =
    μ0Ni
    2R

    ⇒ L =
    N
    =
    μ0Ni
    A
    i2R

    =
    μ0N2
    A ⇒ L ∝ N2
    2R

    Correct Option: C

    L =
    ; φ = BA;
    i

    B =
    μ0Ni
    2R

    ⇒ L =
    N
    =
    μ0Ni
    A
    i2R

    =
    μ0N2
    A ⇒ L ∝ N2
    2R



  1. A rectangular coil of 20 turns and area of cross-section 25 sq. cm has a resistance of 100Ω. If a magnetic field which is perpendicular to the plane of coil changes at a rate of 1000 teals per second, the current in the coil is









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    i =
    e
    =
    nAdB/dt
    RR

    =
    20 × (25 × 10-4) × 1000
    = 0.5 A
    100

    Correct Option: C

    i =
    e
    =
    nAdB/dt
    RR

    =
    20 × (25 × 10-4) × 1000
    = 0.5 A
    100


  1. The total charge induced in a conducting loop when it is moved in a magnetic field depend on









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    q = idt =
    1
    edt
    R

    =
    1
    - dφ
    dt
    Rdt

    =
    1
    R

    (taking only magnitude of e) ​
    Hence, total charge induced in the conducting loop depends upon the total change in magnetic flux.

    Correct Option: C

    q = idt =
    1
    edt
    R

    =
    1
    - dφ
    dt
    Rdt

    =
    1
    R

    (taking only magnitude of e) ​
    Hence, total charge induced in the conducting loop depends upon the total change in magnetic flux.



  1. A 100 millinery coil carries a current of 1A. Energy stored in its magnetic field is​









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    E =
    1
    Li2
    2

    =
    1
    × (100 × 10-3) × 12 = 0.05 J
    2

    Correct Option: C

    E =
    1
    Li2
    2

    =
    1
    × (100 × 10-3) × 12 = 0.05 J
    2