Calculate the emf in a material with flux linkage of 3.5t2 at 2 seconds.
Explanation: The emf induced in a material with flux linkage is given by Vemf = -dλ/dt. On substituting λ= 3.5t2, we get emf = -7t. At time t = 2sec, the emf will be -14 units.
Find the emf induced in a coil of 60 turns with a flux rate of 3 units.
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.
Find the electric field intensity of a charge 2.5C with a force of 3N.
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.
The electric field intensity of a field with velocity 10m/s and flux density of 2.8 units is
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.
The line integral of the electric field intensity is
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.
Which of the following relations is correct?
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.
For static fields, the curl of E will be
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.
The line integral of which parameter is zero for static fields?
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.
The magnitude of the conduction current density for a magnetic field intensity of a vector yi + zj + xk will be
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.
The charge density of a field with a position vector as electric flux density is given by
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.