MCQs on Electromagnetic Theory | Page – 11

  1. Given potential function in free space to be V(x) = 50 (50 x2 + 50 y2 + 50 z2) volts, the magnitude (in volt/m) and the direction of the electric field at a point (1, - 1, 1), where the dimension are in meters are

    Voltage and electric field are related as
    E = - Gradient of V
    - \left[ \frac{ \delta V_x}{ \delta x} u_x + \frac{ \delta v_y} { \delta y} u_y + \frac {\delta V_z}{ \delta z} u_z \rigth ] = - (100x u_x + 100y u_y +100z u_z)
    E (1, - 1, 1) = - [100 u_x – 100 u_y +100 u_z]
    E (1, - 1, 1) = - 100 \sqrt{3} [ \frac{- u_x + u_y - u_z}{ \sqrt{3}}] .

  2. Pointing vector is associated with which of the following?

    P = E × H.

  3. For static magnetic field Maxwell’s curl equation is given by

    For static field, time varying factor will be zero.
    ∇ × H = J + δ D/δ t
    Or, ∇ × H = J
    Or, ∇ × B =μo J.

  4. J = o is frequency known as

    ∇ × J = - δ&rho/δ t
    For steady currents,
    δ&rho/δ t = O.
    ∴ ∇ × J = O is for steady current continuity equation.

  5. A lossy capacitor Cx, rated for operation at 5 KV, 50 Hz is represented by an equivalent circuit with an ideal capacitor Cp in parallel with a resistor Rp. The value of Cp is found to be 0.102 μ F and the value of Rp = 1.25 M ω. Then the power loss and tanδ of the lossy capacitor operating at the rated voltage, respective are

    Power loss = V rated2/Rp = 5 KV/125 Mω
    = 20 W
    For an parallel combination of resistance and capacitor
    Tanδ = 1/(ω Cp Rp).

  6. An inductor designed with 400 turns coil wound on an iron core of 16 cm2 cross sectional area and with a cut of air gap length of 1 mm. The coil is connected to a 230 V, 50 Hz supply. Neglected coil resistance, core loss, iron reluctance and leakage inductance. Find the current in the inductor.

    Inductance is given as
    L = \frac{ \mu_o N^2 A}{l}
    I = \frac{V}{X_L} = \frac{V}{Z \pi f L}
    .

  7. In above questions no. 6, the average force on the core to reduce the air gap will be

    Energy stored in inductor is given as,
    E = (1/2) LI2
    Force required to reduce the air gap of length 1 mm is F = E/l = E/1 mm.

  8. The induction of a long solenoid of length 1000 mm wound uniformly with 3000 turns on a cylindrical paper table of 60 mm diameter is

    Inductor of solenoid is given as
    L = \frac{ \mu_O N^2 A}{l} .

  9. A parallel plate capacitor is field with two dielectrics of ε1 and ε 2 lengthwise equally. The capacitance of the combination is

    Capacitance = ε A/d.
    ε = εo εr
    So,
    [math C_1 = \frac{ \epsilon_0 \epsilon_1 A}{2d} \; \; \; C_2 = \frac{\epsilon_0 \epsilon_2 A}{2d} [/math]
    C = C1 + C2.

  10. 986.8y uz

    μr = χm + 1. Where, χm = magnetic susceptibility H = B/μ.

  11. In above questions 10, the Magnetization M is in KA/m

    M = χ H.

  12. The magnetization curve for an iron alloy is approximately given as
    B= (1/3) H +H2 μ Wb/m2
    If H increase from 0 to 210 A/m, the energy store per unit volume in alloy is

    W = \int_{0}^{H_0} H.dB = \int_{0}^{H_0} H (\frac{1}{3} + 2H) dH .

  13. Three uniform current sheets are located is free space as follows: 8 uz A/m at y = 0, - 4 uz A/m at y = 1 and – 4 uz at y = - 1. Let F be the vector force per meter length exerted on a current filament carrying 7 mA in the uL direction. If the current filament is located at x = 0, y = 0.5 and uL = uz, then force F is

    Within the region – 1 < y < 1, the magnetic field from the two other outer sheets (carrying – 4 uz A/m) cancel, leaving only the field from the centre sheet.
    ∴ H = - 4 ux A/m (0 < y <1) and H = 4 ux A/m (-1 < y <0).
    Outside (y>1 and y < -1) the field from all three sheet cancel leaving H = 0 (y > 1, y < -1)
    So at x = 0, y = 0.5, F/m = I uz × B = (7 × 10-3 uz × - 4 μ0ux.

  14. If magnetization is given by H = (6/a) (- y uz + x uy) in a cube of a size a, the magnetization volume current density is

    J = ∇ × M = 12/a z.

  15. The value of ∫dl along a circle of radius z units is

    Close integral of dl. Is always zero whatever may be the closed path. If is so because dl is an element of displacement (not of simply length) and around a closed path in staring at one point and reaching back over there, net displacement is zero and hence ∫dl = 0.

  16. There concentric spherical shell of radius R1, R2 and R 3 (R1 < R2 < R R3) carry charges – 1, - 2 and 4 coulombs respectively. The charge in coulombs on the inner and outer surface respectively of the outermost shell is

    As far as the conducting spherical shell of radius R3 is concerned, the charge contained in the coulomb enclosed by it = (- 1) + (- 2) = - 3 C. This will induce a charge of + 3c, on the inner surface and a charge of – 3 C on the outer surface of this third shell.
    Also net charge reside on the outer surface has a charge of + 3c and outer surface has 4 – 3 = 1 C.

  17. The intrinsic impedance of copper at high frequency is

    Intrinsic impedance η = &radiac;{ω μ σ}.

  18. Poyniting vector has the dimensions

    Poynting vector represents the directional energy flux density of an electromagnetic field. The unity of poynting vector will be the area. So the dimension of poynting vector is watt per metre square.

  19. A medium is said to be isotropic when

    An isotropic medium is one such that the permittivity, permeability of the medium are uniform in all directions of the medium, the most simple instance being free space.

  20. For incidence from dielectric medium1 (ε1) into dielectric medium2 (ε2) the Brewster angle &theata;p and the corresponding angle of transmission &theata;t for ε21 = 3 will be respectively

    Brewster angle &theata; is given by
    Tan&theata;1 = &radiac; {ε2; 1} = &radiac;3
    ∴ &theata;1 = 60°
    Now sin&theata;1 = &radiac; ε2; sin&theata;2 = ε1
    \frac{ \sin 60^{ \sirc}}{ \sin \theta_2} = \frac{ \epsilon_1}{ \sqrt 3}
    ∴ &theata;2 = 30°.