A parallel-plate capacitor, made of two circular plates of radius R = 10 cm, is connected in series with a resistor of resistance R = 100 Ω

Question

A parallel-plate capacitor, made of two circular plates of radius R = 10 cm, is connected in series with a resistor of resistance R = 100 Ω and a battery of emf E = 10 V. What is the maximum value of the magnetic field induced by the displacement current between the plates, at a distance r = 5 cm from the axis of the plates while the capacitor is being charged?

in progress 0
Thái Dương 10 mins 2021-07-22T16:53:17+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-22T16:54:35+00:00

    Answer:

    Explanation:

    At the time when the capacitor starts charging , current through will be maximum .

    Current in the circuit = 10 / 100 = .1 A .

    This is called displacement current Id = .1 A .

    magnetic field between the plates will be due to this displacement current .

    Magnetic field lines  will be  circular in shape.

    If B be the magnetic field , we shall apply ampere’s circuital law to find magnetic field due to displacement current at a distance of 5 cm from axis.

    ∫ B dl = μ₀ Id

    B x 2π x  .05  = 8.85 x 10⁻¹² x .1

    B = 2.82 x 10⁻¹² T .

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )