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Chapter 30: Q. 74 (page 874)

74. II The inductor in FIGURE P30.74 is a 9.0−cm-long, 2.0−cm−diameter solenoid wrapped with 300 turns. What is the current in the circuit 10μsafter the switch is moved from a to b ?

Short Answer

Expert verified

Hence, the required current is0.093A

Step by step solution

01

Step 1. Given information

The number of turns if 300, and length of inductor is 9 cm

The radius of the coil is,

r=d2r=2cm1m100cm2=0.01m

02

Step 2. Explanation

The inductance of the coil is,

L=μ0N2AlL=μ0N2πr2lL=4π×10−7T⋅m/A(300)2π(0.01m)29cm1m100cm=3.94×10−4H

According to ohm's law, the initial current is,

I0=V0RI0=6V30Ω=0.2A

The expression for the current in the coil is,

I=I0e−tτI=I0e−tLR=I0e−RtLI=I0e−RtLI=(0.2A)e−(30Ω)(10μs)10−4s1/s=0.093A

Hence, the required current is0.093A

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