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Electromagnetic Induction Test - 10

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Electromagnetic Induction Test - 10
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  • Question 1
    1 / -0

    Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimension of permittivity is

    Solution

    Permittivity = [Charge]2 × [Force]1 × [Distance]2
    Or, ε = [IT1]2 × [M1 LT-2]1 × [M0 L1 T0]2 = [M1 L-3 TI2] = M-1L-3T4A2

  • Question 2
    1 / -0

    An ac generator consists of 8 turns of wire, each of area A = 0.0900 m2, and the total resistance of the wire is 12.0Ω. The loop rotates in a 0.500 T magnetic field at a constant frequency of 60.0 Hz. Maximum induced emf is

    Solution

    e = NBAω = NBA2πn = 8 × 0.5 × 0.09 × 2 × 3.14 × 60 = 136V

  • Question 3
    1 / -0

    Two coils are placed closed to each other. The mutual inductance of the pair of coils depends upon:

    Solution

    The mutual inductance of a pair of coils depends upon the relative position and orientation of the two coils.

  • Question 4
    1 / -0

    What happens to the current in coil while accelerating a magnet inside it?

    Solution

    A change in the magnetic field induces an emf. When there is an emf, there has to be current. Hence, when the magnet is moved inside a coil, the current in it increases.

  • Question 5
    1 / -0

    If two coils of inductances L1 and L2 are linked such that their mutual inductance is M, then the maximum value of M is

    Solution

    M = k√L1 L2

    here k is coefficient of coupling. Its maximum value is 1 for tight coupling.

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