A circular coil has a 7.85 cm radius and consists of 36.0 closely wound turns of wire. An externally produced magnetic field of magnitude 2.

Question

A circular coil has a 7.85 cm radius and consists of 36.0 closely wound turns of wire. An externally produced magnetic field of magnitude 2.95 mT is perpendicular to the coil. (a) If no current is in the coil, what magnetic flux links its turns

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Thu Thảo 4 years 2021-07-27T14:45:21+00:00 2 Answers 14 views 0

Answers ( )

  1. Orla
    0
    2021-07-27T14:46:24+00:00
    Reply

    Answer:

    Φ = 2.06×10^-3weber

    Explanation:

    Magnetic flux density is the ratio of the flux links to the area of the coil.

    This can be expressed as

    B =Φ/NA where:

    Φ is the magnetic flux

    B is the magnetic field strength

    A is the area of the coil.

    N is the number of turn in the wire

    Given

    B = 2.95 mT = 2.95×10^-3T

    N = 36

    radius = 7.85cm = 0.0785m

    A = πr² ( since it is a circular coil)

    A = π(0.0785)²

    A = 0.0194m²

    From the formula Φ = BAN

    Φ = 2.95×10^-3 × 0.0194×36

    Φ = 2.06×10^-3weber

  2. nguyetanh
    0
    2021-07-27T14:47:17+00:00
    Reply

    Answer:

    2.056 x 10⁻³ Wb

    Explanation:

    The magnetic flux, Φ₀, through a single turn of a coil is given by the dot product between the magnetic field, B, and the area, A, of the coil. i.e

    Φ₀ = B.A cos θ         ————–(i)

    Where;

    θ = angle between the area vector and the magnetic field.

    But for a coil with N number of turns, the magnetic flux, Φ₀, is actually the product of the number of turns and the magnetic flux through its single turn, Φ₀. i.e

    Φ = N Φ₀     ————–(ii)

    Substitute the value of Φ₀ in equation (i) into equation (ii) as follows;

    Φ = N BAcosθ       ————–(iii)

    From the question, the coil is perpendicular to the magnetic field. This implies that the area vector is parallel to the magnetic field. Therefore equation (iii) becomes;

    Φ = N BA        ——————(iv)

    Also,

    A = πr²

    Where;

    r = radius of the coil

    Substitute A = πr² into equation (iv) as follows;

    Φ = N B(πr²)

    Φ = Nπr²B       ————–(v)

    From the question;

    N = 36.0

    r = 7.85cm = 0.0785m

    B = 2.95mT = 2.95 x 10⁻³ T

    Substitute these values into equation (v) as follows and take π = 3.142;

    Φ = 36.0 x 3.142 x 0.0785² x 2.95 x 10⁻³

    Φ = 0.002056

    Φ = 2.056 x 10⁻³ Wb

    Therefore, the magnetic flux linking its turns is 2.056 x 10⁻³ Wb

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