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The magnetic field at a point inside a 2.0 mH inductor coil becomes 0.80 of its maximum value in `20 mu s ` when the inductor is joined to a battery. Find the resistance of the circuit.

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`l=i_0(1-e^(-t//(tau))`
`(mu_0)ni=(mu_0)ni_(0)(1-e^(-t//(tau))`
`implies B=B_(0)(t-e^(-(tR)//(L))`
`implies0.8 (B_0)=B_(0)(1-e^((-20xx10^(-6)R)/(2xx10^(-3))))`
`impliesB=B_0=(1-e^(-R//(100)))`
`implies (e^(-R//(100)))=0.2`
`impliesIn (e^(-R//(100)))=In(0.2)`
`-(R)/(100)=-1.609`
`implies R=160.9=160 (Omega)`
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HC VERMA-ELECTROMAGNETIC INDUCTION-Exercises
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  7. A solenoid having inductance 4.0 H and resistance 10 Omega is connecte...

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  8. The magnetic field at a point inside a 2.0 mH inductor coil becomes 0...

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  13. A constant current exists in an inductor coil connected to a battery. ...

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  14. Consider the circuit shown in .(a) find the current through the batter...

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  18. The mutual inductance between two coils is 2.5 H. If the current in on...

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