Question:
The magnetic field at a point inside the 2.0 mH inductor coil becomes 0.80 of its maximum value in \(20\,\mu\text{s}\) when the inductor is joined to a battery. Then the resistance of the circuit is:
The magnetic field at a point inside the 2.0 mH inductor coil becomes 0.80 of its maximum value in \(20\,\mu\text{s}\) when the inductor is joined to a battery. Then the resistance of the circuit is:
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Use i = i_max(1 minus e to the minus Rt/L); set it to 0.8 and solve R = (L/t) ln 5.
Updated On: Jul 2, 2026
- 120 ohm
- 440 ohm
- 260 ohm
- 160 ohm
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The Correct Option is D
Solution and Explanation
Step 1: In an L-R circuit the current grows as \(i=i_{\max}\left(1-e^{-Rt/L}\right)\). Since the magnetic field of the coil is proportional to the current, \(B/B_{\max}=i/i_{\max}=0.80\).
Step 2: Set up the equation:
\[0.80=1-e^{-Rt/L}\quad\Rightarrow\quad e^{-Rt/L}=0.20.\]
Step 3: Take the natural logarithm:
\[\frac{Rt}{L}=\ln 5=1.609.\]
Step 4: Solve for R with \(L=2.0\times10^{-3}\,\text{H}\) and \(t=20\times10^{-6}\,\text{s}\):
\[R=\frac{L\ln 5}{t}=\frac{(2.0\times10^{-3})(1.609)}{20\times10^{-6}}=160.9\,\Omega\approx160\,\Omega.\]
This matches option (D).
\[\boxed{R\approx160\,\Omega}\]
Step 2: Set up the equation:
\[0.80=1-e^{-Rt/L}\quad\Rightarrow\quad e^{-Rt/L}=0.20.\]
Step 3: Take the natural logarithm:
\[\frac{Rt}{L}=\ln 5=1.609.\]
Step 4: Solve for R with \(L=2.0\times10^{-3}\,\text{H}\) and \(t=20\times10^{-6}\,\text{s}\):
\[R=\frac{L\ln 5}{t}=\frac{(2.0\times10^{-3})(1.609)}{20\times10^{-6}}=160.9\,\Omega\approx160\,\Omega.\]
This matches option (D).
\[\boxed{R\approx160\,\Omega}\]
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