Electromagnetic theory miscellaneous
- If the vectors → A and →B are conservative then—
-
View Hint View Answer Discuss in Forum
→A and →B both are conservative field vectors.
Therefre, ∇ × →A = 0 and ∇ × →B is also zero.
Now ∇. ( →A × →B) = – ∇. ∇ →A + ∇. ∇ × →B = 0
Hence, ( →A × →B) is solenoidal.Correct Option: A
→A and →B both are conservative field vectors.
Therefre, ∇ × →A = 0 and ∇ × →B is also zero.
Now ∇. ( →A × →B) = – ∇. ∇ →A + ∇. ∇ × →B = 0
Hence, ( →A × →B) is solenoidal.
- If
is the polarization vector and→ n
is the direction of propagation of a plane electromagnetic wave, then—→ k
-
View Hint View Answer Discuss in Forum
The polarization vector and direction of travel are perpendicular to each other. n. k = 0
Correct Option: C
The polarization vector and direction of travel are perpendicular to each other. n. k = 0
- A straight wire of circular cross-section carries a direct current I, as shown in figure below. If R is the resistance per unit length of the wire, then the Poynting vector at the surface of the wire will be—
-
View Hint View Answer Discuss in Forum
R is the resistance per unit length and I is the current flowing. Therefore, power loss occurring in the conductor per unit length is I2R. This power is furnished on E and H fields at the surface of the wire. According to Poynting theorem E × H is the power flow per unit area. Since the power is fed from outside through the cylindrical surface of the conductor, E × H at every point on the surface is radial and directed into the surface. Therefore, Poynting vector is
I2R(−n) = I2R (−n) surface area 2∏r × 1
where n is the unit radial vector directed outward.Correct Option: B
R is the resistance per unit length and I is the current flowing. Therefore, power loss occurring in the conductor per unit length is I2R. This power is furnished on E and H fields at the surface of the wire. According to Poynting theorem E × H is the power flow per unit area. Since the power is fed from outside through the cylindrical surface of the conductor, E × H at every point on the surface is radial and directed into the surface. Therefore, Poynting vector is
I2R(−n) = I2R (−n) surface area 2∏r × 1
where n is the unit radial vector directed outward.
- The velocity of the plane wave esin2(ωt – βx) is—
-
View Hint View Answer Discuss in Forum
The wave is
esin2(ωt − βx) = esin2β 
ω t − x 
t
When the argument is reduced to the form kt – x, the coefficient of t is the velocity of propagation. Hence, the velocity of the wave is ω/β.Correct Option: D
The wave is
esin2(ωt − βx) = esin2β ω t − x t
When the argument is reduced to the form kt – x, the coefficient of t is the velocity of propagation. Hence, the velocity of the wave is ω/β.
- The frequency of the power wave associated with an electromagnetic wave having an E field as
E = e– z/ δ cos (ωt – z/δ),
is given by—
-
View Hint View Answer Discuss in Forum
E = e– z/δ cos (ωt – z/δ)
The radian frequency of E wave is ω, same would be the radian frequency of associated H wave. The frequency of power wave is double the corresponding frequency of E or H wave. Thus the radian frequency is 2 ω and cyclic frequency 2ω/ 2π = ω/ π.Correct Option: D
E = e– z/δ cos (ωt – z/δ)
The radian frequency of E wave is ω, same would be the radian frequency of associated H wave. The frequency of power wave is double the corresponding frequency of E or H wave. Thus the radian frequency is 2 ω and cyclic frequency 2ω/ 2π = ω/ π.
