A plane circular loop of conducting wire of radius with possesses turn is placed in a uniform magnetic field. The direction of the magnetic

Question

A plane circular loop of conducting wire of radius with possesses turn is placed in a uniform magnetic field. The direction of the magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field-strength Bis increased at a constant rate from to in a time interval of 10s. (a) What is the emf generated around the loop? (b) If the electrical resistance of the loop is , what current flows around the loop as the magnetic field is increased?

in progress 0
Khang Minh 6 months 2021-07-19T08:55:38+00:00 1 Answers 54 views 0

Answers ( )

    0
    2021-07-19T08:57:10+00:00

    Complete question;

    A plane circular loop of conducting wire of radius r-10 cm which possesses 15 turns is placed in a uniform magnetic field. The direction of the magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field-strength is increased at a constant rate from IT to 5T in a time interval of 10 s.

    a) What is the emf generated around the loop?

    b) If the electrical resistance of the loop is 15Ω, what current flows around the loop as the magnetic field is increased?

    Answer:

    A) E.M.F generated around loop = 0.1632 Wb/s

    B)Current in loop; I = 0.0109 A

    Explanation:

    A) We are given;

    Initial magnetic field strength;B1 = 1T

    Final magnetic field strength;B2 = 5T

    Number of turns;N = 15 turns

    Radius; r = 10cm = 0.1m

    Angle;θ = 30°

    Time interval; Δt = 10 s

    Now, the formula for magnetic flux is: Φ = NABcosθ

    Where;

    N is number of turns

    A is area = πr²

    B is magnetic field strength

    θ is angle

    So, initial magnetic flux is;

    Φ1 = NA(B1)cosθ

    Plugging in the relevant values to obtain;

    Φ1 = 15*(π*0.1²)(1)cos30

    Φ1 = 0.4081 Wb

    Similarly, final magnetic flux is;

    Φ2 = NA(B2)cosθ

    Plugging in the relevant values to obtain;

    Φ2 = 15*(π*0.1²)(5)cos30

    Φ2 = 2.0405 Wb

    The time rate of change of the flux is;

    dΦ_B/dt = (Φ2 – Φ1)/Δt

    So, dΦ_B/dt = (2.0405 – 0.4081)/10

    dΦ_B/dt = 0.1632 Wb/s

    Thus, the emf generated around the loop is; E = dΦ_B/dt = 0.1632 Wb/s

    B) from Ohm’s law, the current which flows around the loop in response to the emf is given as;

    I = E/R

    We are given R =15Ω

    Thus; I = 0.1632/15

    I = 0.0109 A

    θπΦ

Leave an answer

Browse

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