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Electron moves along positive x directio...

Electron moves along positive x direction with `v = 4 xx 10^(-5)` m/s experiences force of magnitude `3.2 xx 10^(-20)` N at R value of I is:

A

`0.4 A`

B

`40 A`

C

`4 A`

D

`5 A`

Text Solution

Verified by Experts

The correct Answer is:
C
`F = B_(P) + B_(Q)`
`= (mu_(0))/(2pi)(I_(P))/(5) + (mu_(0))/(2pi) (I_(Q))/(2)`
`= (mu_(0))/(2pi)((2.5)/(5) + (I)/(2))`
`= (mu_(0))/(4pi)(I + 1)`
`= 10^(-7)(I + 1)`
Net force on the electron :
`F = B_(q v)sin 90^(@)`
`3.2 xx 10^(-20) = (10^(-7))(I + 1)(1.6 xx 10^(-19))(4 xx 10^(5))`
`2 + 1 = 5`
`I = 4 A`
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