Q 11 :    

A parallel plate capacitor has a capacitance C=200 pF. It is connected to 230 V AC supply with an angular frequency 300 rad/s. The RMS value of conduction current in the circuit and displacement current in the capacitor respectively are            [2024]

  • 1.38μA and 1.38μA

     

  • 14.3μA and 143μA

     

  • 13.8μA and 138μA

     

  • 13.8μA and 13.8μA

     

(4)

 



Q 12 :    

The electric field in an electromagnetic wave is given by E=i^40cosω(t-zc) NC-1.  The magnetic field induction of this wave is (in SI unit):      [2024]

  • B=j^40cosω(t-zc)

     

  • B=k^40ccosω(t-zc)

     

  • B=i^40ccosω(t-zc)

     

  • B=j^40ccosω(t-zc)

     

(4)

E=40cos(ωt-ωcz) NC-1i^

E is along +x direction

v is along +z direction

and Direction of c^ must be along E×B.

So direction of B will be along +y and magnitude of B will be E0c.

B0=E0c=40c



Q 13 :    

The electric field of an electromagnetic wave in free space is

E=57 cos [7.5×106t5×103(3x+4y)](4i^3j^)N/C

The associated magnetic field in Tesla is:          [2025]

  • B=573×108 cos [7.5×106t5×103(3x+4y)](5k^)

     

  • B=573×108 cos [7.5×106t5×103(3x+4y)](k^)

     

  • B=573×108 cos [7.5×106t5×103(3x+4y)](5k^)

     

  • B=573×108 cos [7.5×106t5×103(3x+4y)](k^)

     

(3)

E = CB

B = 57×53×108

3i^+4j^ is direction of propagation

 (4i^3j^)×(k^)=4j^+3i^



Q 14 :    

A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is Ey=9.3 Vm1. Then, the magnetic field vector of the wave at the point is          [2025]

  • Bz=9.3×108T

     

  • Bz=1.55×108T

     

  • Bz=6.2×108T

     

  • Bz=3.1×108T

     

(4)

E = BC

9.3=B×3×108

B=9.33×108=3.1×108T



Q 15 :    

The magnetic field of an E.M. wave is given by

B=(32i^+12j^)30sin[ω(tzc)] (SI Units)

The corresponding electric field in S.I. units is:          [2025]

  • E=(12i^32j^)30c sin[ω(tzc)]

     

  • E=(34i^+14j^)30c cos[ω(tzc)]

     

  • E=(12i^+32j^)30c sin[ω(t+zc)]

     

  • E=(32i^12j^)30c sin[ω(t+zc)]

     

(1)

B=(32i^+12j^)30sin[ω(tzc)]

E×B=c and E0=B0c

Here E=(32(j^)+12i^)

       E0=30c

      E=(12i^32j^) 30c sin [ω(tzc)]



Q 16 :    

A plane electromagnetic wave propagates along the +x direction in free space. The components of the electric field, E and magnetic field, B vectors associated with the wave in Cartesian frame are:          [2025]

  • Ey, Bx

     

  • Ey, Bz

     

  • Ex, By

     

  • Ez, By

     

(2)

Direction of wave propogation E×B



Q 17 :    

If ε0 denotes the permittivity of free space and ΦE is the flux of the electric field through the area bounded by the closed surface, then dimension of (ε0dϕEdt) are that of:          [2025]

  • Electric field

     

  • Electric potential

     

  • Electric charge

     

  • Electric current

     

(4)

We know that formula for displacement current is given by

id=ε0dϕεdt

Dimension of ε0(dϕEdt) will be

Same as electric current

So [ε0dϕEdt]=A



Q 18 :    

A parallel plate capacitor of area A = 16 cm2 and separation between the plates 10 cm, is charged by a DC current. Consider a hypothetical plane surface of area A0=3.2 cm2 inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6 A. At the same instant the displacement current through A0 is _______ mA.          [2025]



(1200)

id=ic

  Total displacement current = 6 A

Current density, Jd=IA=616

  The current through small area i'=Jd×A'

i'=616×3.2=1.12 A=1200 mA



Q 19 :    

A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5 μF. The dielectric constant of the medium between the capacitor plates is 1. It produces an instantaneous displacement current of 0.25 mA in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be _______ Vs1.



(100)

q = CV

Differentiating, i=CdVdt

 dVdt=0.25×1032.5×106=100010=100