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Current in a long current carrying wire ...

Current in a long current carrying wire is `I=2t`
A conducting loop is placed to the right of this wire. Find

a. magnetic flux `phi_B` passing through the loop.
b. induced `emf|e|` prodced in the loop.
c. if total resistance of the loop is `R`, then find induced current `I_("in")` in the loop.

Text Solution

Verified by Experts

Here no conductor is in motion. So we can apply only method -1. Further, magnetic field of straight wire is non uniform. Therefore, magnetic flux can be obtained by integration.

a. At a distance `x` from the staight wire magnetic field is
`B=mu_0/(2pi) I/x` [in `ox` direction]
Let us take a smal strip of width `dx`.
`:.` Area of this strip is
`dS=c(dx)`
Now, `dS` can also be assumed inwards. or, angle between `B` and `dS` may be assumed to be `0^@`
Therefore, small magnetic flux passing through the loop is
`dphi_B=BdS cos0^@`
`mu_0/(2pi) I/x cdx`
Total magnetic flux is
`phi_B=int_(x=a)^(x=a+b) dphi_B`
`=int_a^(a+b)((mu_0IC)/(2pi)) (dx)/x`
`=(mu_0IC)/(2pi) ln ((a+b)/a)`
Substituting the value of `I` we get
`phi_B=(mu_0ct)/pi ln ((a+b)/a)`
b. `|e|=|(dphi_B)/(dt)|=d/(dt)[(mu_0ct)/pi ln ((a+b)/a)]`
`=(mu_0c)/pi ln ((a+b)/a)`
c. Induced current
`I_("in")=(|e|)/R=(mu_0C)/(piR) ln ((a+b)/a)`
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