Question:
A long solenoid having \(100\) turns per cm carries a current of \(\dfrac{4}{\pi}\,\text{A}\). At the centre of it is placed a coil of \(200\) turns of cross-sectional area \(25\,\text{cm}^2\) having its axis parallel to the field produced by the solenoid. When the direction of the current in the solenoid is reversed within \(0.04\,\text{s}\), the induced emf in the coil is:
A long solenoid having \(100\) turns per cm carries a current of \(\dfrac{4}{\pi}\,\text{A}\). At the centre of it is placed a coil of \(200\) turns of cross-sectional area \(25\,\text{cm}^2\) having its axis parallel to the field produced by the solenoid. When the direction of the current in the solenoid is reversed within \(0.04\,\text{s}\), the induced emf in the coil is:
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When current in a solenoid is reversed, the magnetic field changes from \(+B\) to \(-B\). Therefore, the change in magnetic field is \(2B\), not just \(B\).
Updated On: Jun 26, 2026
- \(0.2\,\text{V}\)
- \(0.4\,\text{V}\)
- \(0.002\,\text{V}\)
- \(0.016\,\text{V}\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the magnetic field inside a long solenoid.
The magnetic field inside a long solenoid is \[ B=\mu_0 nI \] where \[ \mu_0=4\pi\times 10^{-7}\,\text{T m A}^{-1} \] \[ n=\text{number of turns per unit length} \] and \[ I=\text{current through the solenoid} \]
Step 2: Convert turns per cm into turns per metre.
Given, \[ 100 \text{ turns per cm} \] Since, \[ 1\,\text{m}=100\,\text{cm} \] therefore, \[ n=100\times 100=10^4\,\text{turns m}^{-1} \]
Step 3: Calculate the magnetic field.
Given current, \[ I=\frac{4}{\pi}\,\text{A} \] So, \[ B=(4\pi\times 10^{-7})(10^4)\left(\frac{4}{\pi}\right) \] \[ B=16\times 10^{-3}\,\text{T} \] \[ B=0.016\,\text{T} \]
Step 4: Find the change in magnetic field.
When the direction of current is reversed, the magnetic field changes from \(+B\) to \(-B\).
Hence, total change in magnetic field is \[ \Delta B=2B \] \[ \Delta B=2(0.016) \] \[ \Delta B=0.032\,\text{T} \]
Step 5: Calculate the induced emf.
Induced emf is given by \[ e=N\frac{\Delta \Phi}{\Delta t} \] Since, \[ \Phi=BA \] therefore, \[ e=N\frac{A\Delta B}{\Delta t} \] Given, \[ N=200 \] \[ A=25\,\text{cm}^2=25\times 10^{-4}\,\text{m}^2 \] \[ A=2.5\times 10^{-3}\,\text{m}^2 \] and \[ \Delta t=0.04\,\text{s} \] Substituting the values, \[ e=200\times \frac{(2.5\times 10^{-3})(0.032)}{0.04} \] \[ e=200\times \frac{8\times 10^{-5}}{0.04} \] \[ e=200\times 2\times 10^{-3} \] \[ e=0.4\,\text{V} \]
Step 6: Final conclusion.
Therefore, the induced emf in the coil is \[ \boxed{0.4\,\text{V}} \]
The magnetic field inside a long solenoid is \[ B=\mu_0 nI \] where \[ \mu_0=4\pi\times 10^{-7}\,\text{T m A}^{-1} \] \[ n=\text{number of turns per unit length} \] and \[ I=\text{current through the solenoid} \]
Step 2: Convert turns per cm into turns per metre.
Given, \[ 100 \text{ turns per cm} \] Since, \[ 1\,\text{m}=100\,\text{cm} \] therefore, \[ n=100\times 100=10^4\,\text{turns m}^{-1} \]
Step 3: Calculate the magnetic field.
Given current, \[ I=\frac{4}{\pi}\,\text{A} \] So, \[ B=(4\pi\times 10^{-7})(10^4)\left(\frac{4}{\pi}\right) \] \[ B=16\times 10^{-3}\,\text{T} \] \[ B=0.016\,\text{T} \]
Step 4: Find the change in magnetic field.
When the direction of current is reversed, the magnetic field changes from \(+B\) to \(-B\).
Hence, total change in magnetic field is \[ \Delta B=2B \] \[ \Delta B=2(0.016) \] \[ \Delta B=0.032\,\text{T} \]
Step 5: Calculate the induced emf.
Induced emf is given by \[ e=N\frac{\Delta \Phi}{\Delta t} \] Since, \[ \Phi=BA \] therefore, \[ e=N\frac{A\Delta B}{\Delta t} \] Given, \[ N=200 \] \[ A=25\,\text{cm}^2=25\times 10^{-4}\,\text{m}^2 \] \[ A=2.5\times 10^{-3}\,\text{m}^2 \] and \[ \Delta t=0.04\,\text{s} \] Substituting the values, \[ e=200\times \frac{(2.5\times 10^{-3})(0.032)}{0.04} \] \[ e=200\times \frac{8\times 10^{-5}}{0.04} \] \[ e=200\times 2\times 10^{-3} \] \[ e=0.4\,\text{V} \]
Step 6: Final conclusion.
Therefore, the induced emf in the coil is \[ \boxed{0.4\,\text{V}} \]
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