UNIT WISE MCQ QUESTION BANK

CLASS: T.Y. B. Tech SUBJECT: EFT

UNIT – 2 Steady Electric Currents and Static Magnetic Fields

Biot Savart Law

1. Biot Savart law in magnetic field is analogous to which law in electric field?
a) Gauss law
b) Faraday law
c) Coulomb’s law
d) Ampere law

Answer: c
Explanation: Biot Savart law states that the magnetic flux density H = I.dl sinθ/4πr2, which is
analogous to the electric field F = q1q2/4πεr2, which is the Coulomb’s law.

2. Which of the following cannot be computed using the Biot Savart law?
a) Magnetic field intensity
b) Magnetic flux density
c) Electric field intensity
d) Permeability

Answer: c
Explanation: The Biot Savart law is used to calculate magnetic field intensity. Using which we can
calculate flux density and permeability by the formula B = μH.

3. Find the magnetic field of a finite current element with 2A current and height 1/2π is
a) 1
b) 2
c) 1/2
d) 1/4

Answer: a
Explanation: The magnetic field due to a finite current element is given by H = I/2πh. Put I = 2
and h = 1/2π, we get H = 1 unit.

4. Calculate the magnetic field at a point on the centre of the circular conductor of radius 2m with
current 8A.
a) 1
b) 2
c) 3
d) 4

Answer: b
Explanation: The magnetic field due to a point in the centre of the circular conductor is given by H
= I/2a. Put I = 8A and a = 2m, we get H = 8/4 = 2 units.
5. The current element of the solenoid of turns 100, length 2m and current 0.5A is given by,
a) 100 dx
b) 200 dx
c) 25 dx
d) 50 dx

Answer: c
Explanation: The current element of the solenoid is given by NI dx/L. Put N = 100, I = 0.5 and L =
2 to get, I dx = 100 x 0.5 x dx/2 = 25 dx.

6. Find the magnetic field intensity at the centre O of a square of the sides equal to 5m and
carrying 10A of current.
a) 1.2
b) 1
c) 1.6
d) 1.8

Answer: d
Explanation: The magnetic field is given by H = 4I/√2πω. Put I = 10 and ω = 5m. Thus H = 4 x
10/√2π(5) = 1.8 unit.

7. Find the magnetic flux density when a point from a finite current length element of current 0.5A
and radius 100nm.
a) 0
b) 0.5
c) 1
d) 2

Answer: c
Explanation: The magnetic flux density is B = μH, where H is given by I/2πr. Put μ = 4π x 10-7, I =
0.5 and r = 10-7, we get B = 4π x 10-7 x 0.5/2π x 10-7 = 1 unit.

8. In a static magnetic field only magnetic dipoles exist. State True/False.

a) True
b) False

Answer: a
Explanation: From Gauss law for magnetic field, we get divergence of the magnetic flux density is
always zero (ie, Div(B) = 0). This implies the non-existence of magnetic monopole.

9. The magnetic field intensity will be zero inside a conductor. State true/false.
a) True
b) False

Answer: b
Explanation: Electric field will be zero inside a conductor and magnetic field will be zero outside
the conductor. In other words, the conductor boundary, E will be maximum and H will be
minimum.
10. Find the magnetic field when a circular conductor of very high radius is subjected to a current
of 12A and the point P is at the centre of the conductor.
a) 1
b) ∞
c) 0
d) -∞

Answer: c
Explanation: The magnetic field of a circular conductor with point on the centre is given by I/2a. If
the radius is assumed to be infinite, then H = 12/2(∞) = 0

Ampere Law
1. The point form of Ampere law is given by
a) Curl(B) = I
b) Curl(D) = J
c) Curl(V) = I
d) Curl(H) = J

Answer: d
Explanation: Ampere law states that the line integral of H about any closed path is exactly equal
to the direct current enclosed by that path. ∫ H.dl = I The point form will be Curl (H) = J.

2. The Ampere law is based on which theorem?

a) Green’s theorem
b) Gauss divergence theorem
c) Stoke’s theorem
d) Maxwell theorem

Answer: c
Explanation: The proof of the Ampere’s circuital law is obtained from Stoke’s theorem for H and J
only.

3. Electric field will be maximum outside the conductor and magnetic field will be maximum inside
the conductor. State True/False.
a) True
b) False

Answer: a
Explanation: At the conductor-free space boundary, electric field will be maximum and magnetic
field will be minimum. This implies electric field is zero inside the conductor and increases as the
radius increases and the magnetic field is zero outside the conductor and decreases as it
approaches the conductor.

4. Find the magnetic flux density of a finite length conductor of radius 12cm and current 3A in air(
in 10-6 order)
a) 4
b) 5
c) 6
d) 7

Answer: b
Explanation: The magnetic field intensity is given by H = I/2πr, where I = 3A and r = 0.12. The
magnetic flux density in air B = μ H, where μ = 4π x 10-7.Thus B = 4π x 10-7 x 3/2π x 0.12 = 5x 10-
6
units.

5. Calculate the magnetic field intensity due to a toroid of turns 50, current 2A and radius 159mm.
a) 50
b) 75
c) 100
d) 200

Answer: c
Explanation: The magnetic field intensity is given by H = NI/2πrm, where N = 50, I = 2A and rm =
1/2π. Thus H = 50 x 2/2π x 0.159 = 100 units.

6. Find the magnetic field intensity due to an infinite sheet of current 5A and charge density of 12j
units in the positive y direction and the z component is above the sheet.
a) -6
b) 12k
c) 60
d) 6

Answer: d
Explanation: The magnetic field intensity when the normal component is above the sheet is Hx =
0.5 K, where K = 12. Thus we get H = 0.5 x 12 = 6 units.

7. Find the magnetic field intensity due to an infinite sheet of current 5A and charge density of 12j
units in the positive y direction and the z component is below the sheet.
a) 6
b) 0
c) -6
d) 60k

Answer: c
Explanation: The magnetic intensity when the normal component is below the sheet is Hy = -0.5
K, where K = 12.Thus we get H = -0.5 x 12 = -6 units.

8. Find the current density on the conductor surface when a magnetic field H = 3cos x i + zcos x j
A/m, for z>0 and zero, otherwise is applied to a perfectly conducting surface in xy plane.
a) cos x i
b) –cos x i
c) cos x j
d) –cos x j

Answer: b
Explanation: By Ampere law, Curl (H) = J. The curl of H will be i(-cos x) – j(0) + k(-z sin x) = -cos
x i – zsin x k. In the xy plane, z = 0. Thus Curl(H) = J = -cos x i.
9. When the rotational path of the magnetic field intensity is zero, then the current in the path will
be
a) 1
b) 0
c) ∞
d) 0.5

Answer: b
Explanation: By Ampere law, Curl(H) = J. The rotational path of H is zero, implies the curl of H is
zero. This shows the current density J is also zero. The current is the product of the current
density and area, which is also zero.

10. Find the magnetic field intensity when the current density is 0.5 units for an area up to 20
units.
a) 10
b) 5
c) 20
d) 40

Answer: a
Explanation: We know that ∫ H.dl = I. By Stoke’s law, we can write Curl(H) = J. In integral form, H
= ∫ J.ds, where J = 0.5 and ds is defined by 20 units. Thus H = 0.5 x 20 = 10 units.

Magnetic Field Density

1. Identify which of the following is the unit of magnetic flux density?
a) Weber
b) Weber/m
c) Tesla
d) Weber-1

Answer: c
Explanation: The unit of magnetic flux density is weber/m 2. It is also called as tesla.

2. The divergence of H will be

a) 1
b) -1
c) ∞
d) 0

Answer: d
Explanation: We know that the divergence of B is zero. Also B = μH. Thus divergence of H is also
zero.

3. Find the flux contained by the material when the flux density is 11.7 Tesla and the area is 2
units.
a) 23.4
b) 12.3
c) 32.4
d) 21.3

Answer: a
Explanation: The total flux is given by φ = ∫ B.ds, where ∫ds is the area. Thus φ = BA. We get φ =
11.7 x 2 = 23.4 units.

4. Find the current when the magnetic field intensity is given by 2L and L varies as 0->1.
a) 2
b) 1
c) 0.5
d) 0

Answer: b
Explanation: From Ampere law, we get ∫ H.dL = I. Put H = 2L and L = 0->1. On integrating H with
respect to L, the current will be 1A.

5. Find the magnetic field intensity when the flux density is 8 x 10-6 Tesla in the medium of air.
a) 6.36
b) 3.66
c) 6.63
d) 3.36

Answer: a
Explanation: We how that, B = μH. To get H = B/μ, put B = 8 x 10 -6 and μ = 4π x 10-7. Thus H = 8
x 10-6/ 4π x 10-7 = 6.36 units.

6. If ∫ H.dL = 0, then which statement will be true?

a) E = -Grad(V)
b) B = -Grad(D)
c) H = -Grad(Vm)
d) D = -Grad(A)

Answer: c
Explanation: The given condition shows that the magnetic field intensity will be the negative
gradient of the magnetic vector potential.

7. Find the magnetic flux density of the material with magnetic vector potential A = y i + z j + x k.
a) i + j + k
b) –i – j – k
c) –i-j
d) –i-k

Answer: b
Explanation: The magnetic flux density is the curl of the magnetic vector potential. B = Curl(A).
Thus Curl(A) = i(-1) – j(1) + k(-1) = -i – j – k. We get B = -i – j – k.

8. Find the magnetic flux density when a flux of 28 units is enclosed in an area of 15cm.
a) 178.33
b) 186.67
c) 192.67
d) 124.33

Answer: b
Explanation: The total flux is the product of the magnetic flux density and the area. Total flux = B
x A. To get B, put flux/area. B = 28/0.15 = 186.67 units.

9. Find the magnetic flux density B when E is given by 3sin y i + 4cos z j + ex k.

a) ∫(4sin z i – ex j – 3cos y k)dt
b) -∫(4sin z i – ex j – 3cos y k)dt
c) ∫(4sin y i – ex j + 3cos y k)dt
d) -∫(4sin y i + ex j + 3cos y k)dt

Answer: b
Explanation: We know that Curl (E) = -dB/dt. The curl of E is (4sin z i – ex j – 3cos y k). To get B,
integrate the -curl(E) with respect to time to get B = -∫(4sin z i – ex j – 3cos y k)dt.

10. Find current density J when B = 50 x 10-6 units and area dS is 4 units.
a) 9.94
b) 8.97
c) 7.92
d) 10.21

Answer: a
Explanation: To get H, H = B/μ = 50 x 10-6/ 4π x 10-7 = 39.78 units. Also H = ∫ J.dS, where H =
39.78 and ∫ dS = 4. Thus J = 39.78/4 = 9.94 units.

Magnetic Vector Potential

1. The magnetic vector potential is a scalar quantity.
a) True
b) False

Answer: b
Explanation: The magnetic vector potential could be learnt as a scalar. But it is actually a vector
quantity, which means it has both magnitude and direction.

2. Find the magnetic field intensity when the magnetic vector potential x i + 2y j + 3z k.
a) 6
b) -6
c) 0
d) 1

Answer: b
Explanation: The magnetic field intensity is given by H = -Grad(Vm). The gradient of Vm is 1 + 2
+ 3 = 6. Thus H = -6 units.

3. The value of ∫ H.dL will be

a) J
b) I
c) B
d) H

Answer: b
Explanation: By Stoke’s theorem, ∫ H.dL = ∫ Curl(H).dS and from Ampere’s law, Curl(H) = J. Thus
∫ H.dL = ∫ J.dS which is nothing but current I.

4. Given the vector potential is 16 – 12sin y j. Find the field intensity at the origin.
a) 28
b) 16
c) 12
d) 4

Answer: c
Explanation: The field intensity is given by H = – Grad(V). The gradient is given by 0 – 12cos y. At
the origin, the gradient will be -12 cos 0 = -12. Thus the field intensity will be 12 units.

5. Find the vector potential when the field intensity 60x2 varies from (0,0,0) to (1,0,0).
a) 120
b) -20
c) -180
d) 60

Answer: b
Explanation: The field intensity H = -Grad(V). To get V, integrate H with respect to the variable.
Thus V = -∫H.dl = -∫60x2 dx = -20x3 as x = 0->1 to get -20.

6. Find the flux density B when the potential is given by x i + y j + z k in air.

a) 12π x 10-7
b) -12π x 10-7
c) 6π x 10-7
d) -6π x 10-7

Answer: b
Explanation: The field intensity H = -Grad(V). Since the given potential is a position vector, the
gradient will be 3 and H = -3. Thus the flux density B = μH = 4π x 10-7 x (-3) = -12π x 10-7 units.

7. The Laplacian of the magnetic vector potential will be

a) –μ J
b) – μ I
c) –μ B
d) –μ H

Answer: a
Explanation: The Laplacian of the magnetic vector potential is given by Del2(A) = -μ J, where μ is
the permeability and J is the current density.

8. The magnetic vector potential for a line current will be inversely proportional to
a) dL
b) I
c) J
d) R

Answer: d
Explanation: The magnetic vector potential for the line integral will be A = ∫ μIdL/4πR. It is clear
that the potential is inversely proportional to the distance or radius R.

9. The current element of the magnetic vector potential for a surface current will be
a) J dS
b) I dL
c) K dS
d) J dV

Answer: c
Explanation: The magnetic vector potential for the surface integral is given by A = ∫ μKdS/4πR. It
is clear that the current element is K dS.

10. The relation between flux density and vector potential is

a) B = Curl(A)
b) A = Curl(B)
c) B = Div(A)
d) A = Div(B)

Answer: a
Explanation: The magnetic flux density B can be expressed as the space derivative of the
magnetic vector potential A. Thus B = Curl(A).

Magnostatic Energy
1. Find the induced EMF in an inductor of 2mH and the current rate is 2000 units.
a) 4
b) -4
c) 1
d) -1

Answer: b
Explanation: The induced emf is given by e = -Ldi/dt. Put L = 2 x 10-3 and di/dt = 2000 in the
equation. We get e = -2 x 10-3 x 2000 = -4 units.

2. Find the work done in an inductor of 4H when a current 8A is passed through it?
a) 256
b) 128
c) 64
d) 512

Answer: b
Explanation: The work done in the inductor will be W = 0.5 x LI 2. On substituting L = 4 and I = 8,
we get, W = 0.5 x 4 x 82 = 128 units.
3. Find the inductance of a material with 100 turns, area 12 units and current of 2A in air.
a) 0.75mH
b) 7.5mH
c) 75mH
d) 753mH

Answer: a
Explanation: The inductance of any material(coil) is given by L = μ N2A/I. On substituting N = 100,
A = 0.12 and I = 2, we get L = 4π x 10-7 x 1002 x 0.12/2 = 0.75 units.

4. Calculate the magnetic energy when the magnetic intensity in air is given as 14.2 units(in 10 -
4
order)
a) 1.26
b) 2.61
c) 6.12
d) 1.62

5. Calculate the magnetic energy when the magnetic flux density is given by 32 units(in 108 order)
a) 4.07
b) 7.4
c) 0.47
d) 7.04

Answer: a
Explanation: The magnetic energy is given by E = 0.5 μ H2 and we know that μH = B. On
substituting we get a formula E = 0.5 B2/μ. Put B = 32 and in air μ = 4π x 10-7, we get E = 0.5 x
322/4π x 10-7 = 4.07 x 108 units.

6. Calculate the energy when the magnetic intensity and magnetic flux density are 15 and 65
respectively.
a) 755
b) 487.5
c) 922
d) 645

Answer: b
Explanation: The magnetic energy can also be written as E = 0.5 μH2 = 0.5 BH, since B = μH. On
substituting B = 65 and H = 15 we get E = 0.5 x 65 x 15 = 487.5 units.

7. Find the inductance when the energy is given by 2 units with a current of 16A.
a) 15.6mH
b) 16.5mH
c) 16.8mH
d) 15.8mH

Answer: a
Explanation: The energy stored in an inductor is given by E = 0.5 LI2. To get L, put E = 2 and I =
16 and thus L = 2E/I2 = 2 x 2/162 = 15.6mH.
8. Find the power of an inductor of 5H and current 4.5A after 2 seconds.
a) 25.31
b) 50.62
c) 102.4
d) 204.8

Answer: a
Explanation: The energy stored in an inductor is given by E = 0.5 LI2. Thus, put L = 5 and I = 4.5
and we get E = 0.5 x 5 x 4.52 = 50.625 units To get power P = E/t = 50.625/2 = 25.31 units.

9. Find the turns in an solenoid of inductance 23.4mH , current 2A and area 15cm.
a) 900
b) 400
c) 498
d) 658

Answer: c
Explanation: The inductance of any material(coil) is given by L = μ N2A/I.
Put L = 23.4 x 10-3, I = 2 and A = 0.15, we get N as 498 turns.

10. The energy of a coil depends on the turns. State True/False.

a) True
b) False

Answer: a
Explanation: The inductance is directly proportional to square of the turns. Since the energy is
directly proportional to the inductance, we can say both are dependent on each other.

Inductances
1. Calculate the emf of a coil with turns 100 and flux rate 5 units.
a) 20
b) -20
c) 500
d) -500

Answer: d
Explanation: The emf is the product of the turns of the coil and the flux rate. Thus e = -N dφ/dt,
where the negative sign indicates that the emf induced is opposing the flux. Thus e = -100 x 5 = -
500 units.

2. The equivalent inductances of two coils 2H and 5H in series aiding flux with mutual inductance
of 3H is
a) 10
b) 30
c) 1
d) 13
Answer: d
Explanation: The equivalent inductance of two coils in series is given by L = L1 + L2 + 2M, where
L1 and L2 are the self inductances and M is the mutual inductance. Thus L = 2 + 5 + 2(3) = 13H.

3. The expression for the inductance in terms of turns, flux and current is given by
a) L = N dφ/di
b) L = -N dφ/di
c) L = Niφ
d) L = Nφ/i

Answer: a
Explanation: We know that e = -N dφ/dt and also e = -L di/dt. On equating both we get, L =
Ndφ/di is the expression for inductance.

4. The equivalent inductance of two coils with series opposing flux having inductances 7H and 2H
with a mutual inductance of 1H.
a) 10
b) 7
c) 11
d) 13

Answer: b
Explanation: The equivalent inductance of two coils in series with opposing flux is L = L1 + L2 –
2M, where L1 and L2 are the self inductances and M is the mutual inductance. Thus L = 7 + 2 –
2(1) = 7H.

5. A coil is said to be loosely coupled with which of the following conditions?

a) K>1
b) K<1
c) K>0.5
d) K<0.5

Answer: d
Explanation: k is the coefficient of coupling. It lies between 0 and 1. For loosely coupled coil, the
coefficient of coupling will be very less. Thus the condition K<0.5 is true.

6. With unity coupling, the mutual inductance will be

a) L1 x L2
b) L1/L2
c) √(L1 x L2)
d) L2/L1

Answer: c
Explanation: The expression for mutual inductance is given by M = k √(L1 x L2), where k is the
coefficient of coupling. For unity coupling, k = 1, then M = √(L1 x L2).

7. The inductance is proportional to the ratio of flux to current. State True/False.

a) True
b) False

Answer: a
Explanation: The expression is given by L = Ndφ/di. It can be seen that L is proportional to the
ratio of flux to current. Thus the statement is true.

8. Calculate the mutual inductance of two tightly coupled coils with inductances 49H and 9H.
a) 21
b) 58
c) 40
d) 49/9

Answer: a
Explanation: For tightly coupled coils, the coefficient of coupling is unity. Then the mutual
inductance will be M = √(L1 x L2)= √(49 x 9) = 21 units.

9. Find the inductance of a coil with turns 50, flux 3 units and a current of 0.5A
a) 150
b) 300
c) 450
d) 75

Answer: b
Explanation: The self inductance of a coil is given by L = Nφ/I, where N = 50, φ = 3 and I = 0.5.
Thus L = 50 x 3/0.5 = 300 units.

10. The inductance of a coaxial cable with inner radius a and outer radius b, from a distance d, is
given by
a) L = μd ln(b/a)/2π
b) L = 2π μd ln(b/a)
c) L = πd/ln(b/a)
d) L = 0

Answer: a
Explanation: The inductance of a coaxial cable with inner radius a and outer radius b, from a
distance d, is a standard formula derived from the definition of the inductance. This is given by L
= μd ln(b/a)/2π.

Space for rough work