Electromagnetic theory miscellaneous
- The velocity of electromagnetic waves in free space—
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The velocity of electromagnetic waves in free space is independent of frequency.
Correct Option: C
The velocity of electromagnetic waves in free space is independent of frequency.
- The value of ∮ di along a circle of radius 2 units is— ƒ
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∮ dl is always zero whatever may be the closed path it is so because dl is an element of displacement displacement (not of simply length) and around a closed path is starting at one point and reaching back over there, net displacement is zero and hence ∮ dI = 0.
Correct Option: A
∮ dl is always zero whatever may be the closed path it is so because dl is an element of displacement displacement (not of simply length) and around a closed path is starting at one point and reaching back over there, net displacement is zero and hence ∮ dI = 0.
- If the vectors → A and →B are conservative then—
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→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
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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—
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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.