Electromagnetic Theory MCQ || Electromagnetic Theory Questions and Answers

Answer.3. 2 V/m

Explanation

Dielectric constant (ε_r) is defined as the ratio of the electric permittivity of the material to the electric permittivity of free space.

The presence of the dielectric reduces the effective electric field.

The relative permittivity of the vacuum is 1.

As the relative permittivity increases by a factor of 2, the electric field decreases by a factor of 2.

Therefore, the electric field = 2 V/m

Answer.2. 0.6 m

Explanation

Consider new charge + 4μC is placed d m apart from old +4 μC charge and (x + 0.6) m apart from -16 μC charge.

Let,
q_A = + 4 μC at point A
q_B = – 16 μC at point B
q_C = + 6 μC at point C

Since, net force on charge q_C will zero.

∴ |F_CA| = |F_CB|

According to Coulomb’s law the net charge

$F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}$

Where

K is a constant = 9 × 10⁹ Nm²/C²

$\begin{array}{l} \frac{{K{q_A}{q_C}}}{{{d^2}}} = \frac{{K{q_B}{q_C}}}{{{{(0.6 + d)}^2}}}\\ \\ \frac{{24}}{{{d^2}}} = \frac{{96}}{{{{(0.6 + d)}^2}}} \end{array}$

4d² = (0.6 + d)²

4d² = 0.36 + d² + 1.2d

3d² – 1.2d – 0.36 = 0

d₁ = – 0.2 m

d₂ = + 0.6 m

Hence, according to option + 0.6 m distance should a third charge of + 6 μC be placed from + 4 μC so that no force exerts on it will be zero.

Answer.1. Diamagnetic material

Explanation

In diamagnetic materials, the magnetic field in the material is weakened by induced magnetization. Ferromagnetic, ferrimagnetic materials possess permanent magnetization even without an external magnetic field.

Answer.3. 14.4 V

Explanation

Average induced emf is given by

$E = N\frac{{d\phi }}{{dt}}$

Where N is the number of turns

dϕ is changing in flux

dt is changing in time

Calculation:

Given that, number of turns (N) = 360

Change in time (dt) = 0.01 s

Magnetic flux (ϕ) = 200 μWb

Since the flux is reversed, it changes from 200 μWb to -200 μWb, which is a change of 200 – (-200), i.e. 400 μWb

Change in magnetic flux (dϕ) = 400 μWb

$E = 360 \times \frac{{400 \times {{10}^{ – 6}}}}{{0.01}} = 14.4\:V$

Answer.2. 0.04 mWb

Explanation

Mutual inductance between two coils can be written as

M = N₁φ₁₂/I₁

Flux produced in coil X (ϕ1) = 0.05 mWb

As we are just required to find the flux linked with the second coil, we are given that 80% of the flux produced by one coil links with the other.

∴ Flux linked with Y (ϕ12) = 80% of flux produced in coil 1

= 0.05 × 0.8 mWb

0.04 mWb

Answer.1. 0.45 N/m

Explanation

We know the force per unit length between the two conductors is

$\frac{F}{l} = \frac{{{\mu _0}{I_1}{I_2}}}{{2\pi d}}$

Here d = distance = 10 × 10^-3 meter

μ0 = 4π × 10^-7F/m

And currents I₁ = I₂ = 150 Ampere

$\therefore \frac{F}{l} = \frac{{4\pi \times {{10}^{ – 7}} \times 150 \times 150}}{{2\pi \times 10 \times {{10}^{ – 3}}}} = 0.45\:N/m$

Answer.1. Zero

Explanation

The magnetic flux is defined as the number of magnetic field lines passing through a closed surface

Flux can be expressed as

φ =  BACosθ

ϕ  is the magnetic flux

B is the magnetic flux density

A is the area

θ is the angle between the surface area vector of a surface and magnetic field

Given that, θ = 90°

φ =  BACos90° = 0

Answer.1. (1 + M)/H

Explanation

The magnetic flux density is given by

B = μ₀ (H + M)

Where, M = magnetization of the medium and H = magnetic field strength

μ_r is the relative permeability of a medium

⇒ μ₀μ_rH = μ₀ (H + M)

⇒ μ_rH = H + M

⇒ (μ_r – 1) H = M

⇒ (μ_r – 1)  = M/H

μ_r = (1 + M)/H

Answer.3. 10^-3 Wb/m²

Explanation

Given:

r = 5 cm = 5 × 10^-2 m

I = 250 A

We know that Magnetic flux density B

B = μ₀ H

Where

B = Magnetic flux density (Wb/m²)

μ₀ = Absolute permeability = 4π × 10^-7 H/m

$\begin{array}{l} B = {\mu _0}H = {\mu _0}\frac{I}{{2\pi r}}\\ \\ = 4\pi \times {10^{ – 7}} \times \:\frac{{250}}{{2\pi \times 5 \times {{10}^{ – 2}}}} = {10^{ – 3}} \end{array}$

∴ B = 10^-3 Wb/m²

Answer.2. Same as that on the surface

Explanation

Given:

• The electric field inside a conducting sphere is zero, so the potential remains constant at the value it reaches the surface.
• When a conductor is at equilibrium, the electric field inside it is constrained to be zero.
• Since the electric field is equal to the rate of change of potential, this implies that the voltage inside a conductor at equilibrium is constrained to be constant at the value it reaches the surface of the conductor.
• A good example is the charged conducting sphere, but the principle applies to all conductors at equilibrium.