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QUESTION: 1

Calculate the emf in a material with flux linkage of 3.5t^{2} at 2 seconds.

Solution:

Answer: c

Explanation: The emf induced in a material with flux linkage is given by Vemf = -dλ/dt. On substituting λ= 3.5t^{2}, we get emf = -7t. At time t = 2sec, the emf will be -14 units.

QUESTION: 2

Find the emf induced in a coil of 60 turns with a flux rate of 3 units.

Solution:

Answer: b

Explanation: The emf induced is the product of the turns and the flux rate. Thus Vemf = -Ndφ/dt. On substituting N = 60 and dφ/dt = 3, we get emf as -60 x 3 = -180 units.

QUESTION: 3

Find the electric field intensity of a charge 2.5C with a force of 3N.

Solution:

Answer: d

Explanation: The electric field intensity is the electric force per unit charge. It is given by E = F/q. On substituting F = 2.5 and q = 3, we get E = 3/2.5 units.

QUESTION: 4

The electric field intensity of a field with velocity 10m/s and flux density of 2.8 units is

Solution:

Answer: b

Explanation: The electric field is the product of the velocity and the magnetic flux density given by E = v x B. On substituting v = 10 and B = 2.8, we get E = 10 x 2.8 = 28 units.

QUESTION: 5

The line integral of the electric field intensity is

Solution:

Answer: b

Explanation: From the Maxwell first law, the transformer emf is given by the line integral of the electric field intensity. Thus the emf is given by ∫ E.dl.

QUESTION: 6

Which of the following relations is correct?

Solution:

Answer: c

Explanation: The emf induced in a material is given by the line integral of the electric field intensity. Thus EMF = ∫ E.dl is the correct relation.

QUESTION: 7

For static fields, the curl of E will be

Solution:

Answer: b

Explanation: For static fields, the charges will be constant and the field is constant. Thus curl of the electric field intensity will be zero. This implies the field is irrotational.

QUESTION: 8

The line integral of which parameter is zero for static fields?

Solution:

Answer: a

Explanation: The field is irrotational for static fields. Thus curl of E is zero. From Stokes theorem, the line integral of E is same as the surface integral of the curl of E. Since it is zero, the line integral of E will also be zero.

QUESTION: 9

The magnitude of the conduction current density for a magnetic field intensity of a vector yi + zj + xk will be

Solution:

Answer: b

Explanation: From the Ampere circuital law, the curl of H is the conduction current density. The curl of H = yi + zj + xk is –i – j – k. Thus conduction current density is –i – j – k. The magnitude will be √(1 + 1 + 1) = √3 = 1.732 units.

QUESTION: 10

The charge density of a field with a position vector as electric flux density is given by

Solution:

Answer: d

Explanation: The Gauss law for electric field states that the divergence of the electric flux density is the charge density. Thus Div(D) = ρ. For D as a position vector, the divergence of the position vector D will be always 3. Thus the charge density is also 3.

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