Question:
A circular coil has \(100\) turns, radius \(3\;cm\) and resistance \(4\;\Omega\). This coil is co-axial with a solenoid of \(200\) turns/cm and diameter \(4\;cm\). If the solenoid current is decreased from \(2\;A\) to zero in \(0.04\;s\), then the current induced in the coil is
A circular coil has \(100\) turns, radius \(3\;cm\) and resistance \(4\;\Omega\). This coil is co-axial with a solenoid of \(200\) turns/cm and diameter \(4\;cm\). If the solenoid current is decreased from \(2\;A\) to zero in \(0.04\;s\), then the current induced in the coil is
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When a coil surrounds a solenoid, use only the cross-sectional area of the solenoid for magnetic flux, because the magnetic field is mainly confined inside the solenoid.
Updated On: Jun 22, 2026
- \(4\pi^2\;mA\)
- \(8\pi\;mA\)
- \(30.3\;mA\)
- \(45.5\;mA\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the magnetic field inside the solenoid.
Magnetic field inside a long solenoid is \[ B=\mu_0 nI \] Number of turns per unit length is \[ n=200\;\text{turns/cm} \] \[ n=200\times 100=20000\;\text{turns/m} \]
Step 2: Use the area of the solenoid.
Since the coil radius is \(3\;cm\) and solenoid radius is \(2\;cm\), magnetic flux is only through the cross-sectional area of the solenoid.
Radius of solenoid is \[ r=2\;cm=2\times10^{-2}\;m \] So, \[ A=\pi r^2 \] \[ A=\pi(2\times10^{-2})^2 \] \[ A=4\pi\times10^{-4}\;m^2 \]
Step 3: Find rate of change of current.
Current decreases from \(2\;A\) to \(0\;A\) in \(0.04\;s\).
Thus, \[ \left|\frac{dI}{dt}\right|=\frac{2}{0.04} \] \[ =50\;A s^{-1} \]
Step 4: Find induced emf in the coil.
Induced emf is \[ e=N A\mu_0 n\left|\frac{dI}{dt}\right| \] Here, \[ N=100 \] So, \[ e=100(4\pi\times10^{-4})(4\pi\times10^{-7})(20000)(50) \] \[ e=1.6\pi^2\times10^{-2}\;V \]
Step 5: Find induced current.
Using Ohm's law, \[ i=\frac{e}{R} \] \[ i=\frac{1.6\pi^2\times10^{-2}}{4} \] \[ i=0.4\pi^2\times10^{-2}\;A \] \[ i=4\pi^2\times10^{-3}\;A \] \[ i=4\pi^2\;mA \]
Step 6: Final conclusion.
Hence, the induced current in the coil is \[ \boxed{4\pi^2\;mA} \]
Magnetic field inside a long solenoid is \[ B=\mu_0 nI \] Number of turns per unit length is \[ n=200\;\text{turns/cm} \] \[ n=200\times 100=20000\;\text{turns/m} \]
Step 2: Use the area of the solenoid.
Since the coil radius is \(3\;cm\) and solenoid radius is \(2\;cm\), magnetic flux is only through the cross-sectional area of the solenoid.
Radius of solenoid is \[ r=2\;cm=2\times10^{-2}\;m \] So, \[ A=\pi r^2 \] \[ A=\pi(2\times10^{-2})^2 \] \[ A=4\pi\times10^{-4}\;m^2 \]
Step 3: Find rate of change of current.
Current decreases from \(2\;A\) to \(0\;A\) in \(0.04\;s\).
Thus, \[ \left|\frac{dI}{dt}\right|=\frac{2}{0.04} \] \[ =50\;A s^{-1} \]
Step 4: Find induced emf in the coil.
Induced emf is \[ e=N A\mu_0 n\left|\frac{dI}{dt}\right| \] Here, \[ N=100 \] So, \[ e=100(4\pi\times10^{-4})(4\pi\times10^{-7})(20000)(50) \] \[ e=1.6\pi^2\times10^{-2}\;V \]
Step 5: Find induced current.
Using Ohm's law, \[ i=\frac{e}{R} \] \[ i=\frac{1.6\pi^2\times10^{-2}}{4} \] \[ i=0.4\pi^2\times10^{-2}\;A \] \[ i=4\pi^2\times10^{-3}\;A \] \[ i=4\pi^2\;mA \]
Step 6: Final conclusion.
Hence, the induced current in the coil is \[ \boxed{4\pi^2\;mA} \]
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