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The magnetic field of an electromagnetic...

The magnetic field of an electromagnetic wave is given by:
`oversetB=1.6xx10^(-6)cos(2xx10^(7)z+6xx10^(15)t)(2hati+hatj)(Wb)/m^(2)`
The associated electric field will be:

A

` vec E = 4.8 xx 10 ^( 2) cos ( 2 xx 10 ^( 7) z + 6 xx 10 ^(15) t ) ( hati - 2 hatj ) (V)/( m ) `

B

`vecE = 4.8 xx 10 ^( 2 ) cos ( 2 xx 10 ^( 7) z + 6 xx 10 ^( 15 ) t ) ( - hati + 2 hatj ) (V)/(m) `

C

` vec E = 4.8 xx 10 ^2 cos ( 2 xx 10^7 z - 6 xx 10 ^(15) t ) ( - 2 hatj + hati ) (V)/(m) `

D

` vecE = 4.8 xx 10 ^(2) cos (2 xx 10 ^(7) z - 6 xx 10 ^( 15) t ) ( 2 hati + j ) (V)/(m) `

Text Solution

Verified by Experts

`vec B = 1.6 xx 10 ^( -6 ) cos (2 xx 10 ^(7) Z + 6 xx 10 ^( 15) t) (2hati + hatj ) (Wb)/(m ^ 2 ) `
` E _ 0 = cB_ 0 `
` = 3 xx 10 ^( 8) xx 1.6 xx 10 ^( -6 ) = 4.8 xx 10 ^( 2 ) `
From the equation of ` vec B ` we can conclude that the direction of wave is ` - hatk `
Direction of ` vecE = ` direction of ` vecB xx vec v `
` ( 2 hati + hatj ) xx ( - hatk ) = - hati + 2 hatj `
Hence
` vec E = 4.8 xx 10 ^(2) cos ( 2 xx 10 ^( 7) z + 6 xx 10 ^( 15) t ) ( - hati + 2 hatj ) V//m `
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