Electrostatics 7 Question 1

1. The magnetic field of a plane electromagnetic wave is given by

B=B0[cos(kzωt)]i^+B1cos(kz+ωt)j^

where, B0=3×105T and B1=2×106T. The rms value of the force experienced by a stationary charge Q=104C at z=0 is closest to

(Main 2019, 9 April I)

(a) 0.1N

(b) 3×102N

(c) 0.6N

(d) 0.9N

Show Answer

Solution:

  1. Given, magnetic field of an electromagnetic wave is

B=B0[cos(kzωt]i^+B1[cos(kz+ωt]j^

Here, B0=3×105T and B1=2×106T

Also, stationary charge, Q=104C at z=0

As charge is released from the rest at z=0, in this condition.

Maximum electric field, E0=cB0 and E1=cB1

So, E0=c×3×105 and E1=c×2×106

Now,the direction of electric field of an electromagnetic wave is perpendicular to B and to the direction of propagation of wave (E×B) which is k^

So, for E0,E0×B0=k^E0×i^=k^E0=j^

Similarly,

 for E1E1×B1=kE1×j^=k^E1=i^E0=c×3×105(j^)NC1E=c×2×106(+i^)NC1

Maximum force experienced by stationary charge is

Fmax=QE=Q(E0+E1)=Q×c[3×105j^+2×106i^]

|Fmax|=104×3×108×(3×105)2+(2×106)2=3×104×106900+4=3×102×9040.9N

rms value of experienced force is

Frms=Fmax2=0.92=0.707×0.9=0.6363N0.6N



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक