The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression `vec(E) (x,y) = 10 hat(j) cos [(6 x + 8z)]` The magnetic field `vec(B) (x, z, t)` is given by: (c is the velocity of light)
A
`(1)/(c) (6hatk+8hati)cos[(6x-8z+10ct)]`
B
`(1)/(c)(6hatk+8hati)cos[(6x+8z-10ct)]`
C
`(1)/(c)(6hatk-8hati)cos[(6x+8z-10ct)]`
D
`(1)/(c)(6hatk-8hati)cos[(6x+8z+10ct)]`
Text Solution
Verified by Experts
The correct Answer is:
C
`vecE=E_(0)(veck*vecr-omegat)` `vecE(x,y)=10hatjcos(6x+8z)=10j cos[(6hati+8hatk)*(x hati+z hatk)]` `veck=6 hati+8hatk, omega=|veck|c=10c` `vecE(x,y,z,t)=10hatj cos[(6hati+8hatk)*(xhati+z hatk)-10ct]` Now, `vecE xx vecB= (E^(2))/(c)hatk=(E^(2))/(10c)(6hati+8hatk)` `10hatj xx (B_(x)hati+B_(y)hatj+B_(z)hatk)=(10)/(c) (6hati+8hatk)` Looking at the options we can say `vecB=(1)/(c)(6 hatk-8hati)` `vecB=(x,y,z,t)=(1)/(c)(6hatk-8hati)cos[(6x+8z)-10ct]`
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VMC MODULES ENGLISH-JEE MAIN REVISION TEST-3 -PHYSICS (SECTION 1)
