Question:
A particle of mass \(2.2\times 10^{-30}\,\text{kg}\) and charge \(1.6\times 10^{-19}\,\text{C}\) is moving at a speed of \(10\,\text{km s}^{-1}\) in a circular path of radius \(2.8\,\text{cm}\) inside a solenoid. The solenoid has \(25\,\text{turns/cm}\) and its magnetic field is perpendicular to the plane of the particle's path. The current in the solenoid is
Take \(\mu_0=4\pi\times 10^{-7}\,\text{H m}^{-1}\).
A particle of mass \(2.2\times 10^{-30}\,\text{kg}\) and charge \(1.6\times 10^{-19}\,\text{C}\) is moving at a speed of \(10\,\text{km s}^{-1}\) in a circular path of radius \(2.8\,\text{cm}\) inside a solenoid. The solenoid has \(25\,\text{turns/cm}\) and its magnetic field is perpendicular to the plane of the particle's path. The current in the solenoid is
Take \(\mu_0=4\pi\times 10^{-7}\,\text{H m}^{-1}\).
Take \(\mu_0=4\pi\times 10^{-7}\,\text{H m}^{-1}\).
Show Hint
For a charged particle moving in a circular path in a magnetic field, use
\[
qvB=\frac{mv^2}{r}
\]
and for a solenoid, use
\[
B=\mu_0 nI
\]
where \(n\) must be in turns per metre.
Updated On: Jun 22, 2026
- \(1.25\,\text{mA}\)
- \(10.20\,\text{mA}\)
- \(2.50\,\text{mA}\)
- \(1.56\,\text{mA}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use magnetic force as centripetal force.
Since the charged particle moves in a circular path inside the magnetic field, the magnetic force provides the centripetal force.
Therefore, \[ qvB=\frac{mv^2}{r} \] So, \[ B=\frac{mv}{qr} \]
Step 2: Convert the given quantities into SI units.
Given, \[ m=2.2\times 10^{-30}\,\text{kg} \] \[ q=1.6\times 10^{-19}\,\text{C} \] \[ v=10\,\text{km s}^{-1}=10\times 10^3=10^4\,\text{m s}^{-1} \] \[ r=2.8\,\text{cm}=2.8\times 10^{-2}\,\text{m} \]
Step 3: Calculate the magnetic field.
\[ B=\frac{(2.2\times 10^{-30})(10^4)} {(1.6\times 10^{-19})(2.8\times 10^{-2})} \] \[ B=\frac{2.2\times 10^{-26}} {4.48\times 10^{-21}} \] \[ B=4.91\times 10^{-6}\,\text{T} \]
Step 4: Use magnetic field inside a solenoid.
For a long solenoid, \[ B=\mu_0 n I \] So, \[ I=\frac{B}{\mu_0 n} \] The number of turns per unit length is \[ n=25\,\text{turns/cm} \] Since, \[ 1\,\text{cm}=10^{-2}\,\text{m} \] we get \[ n=25\times 100=2500\,\text{turns/m} \]
Step 5: Calculate the current.
\[ I=\frac{4.91\times 10^{-6}} {(4\pi\times 10^{-7})(2500)} \] Using \[ \pi \approx 3.14 \] \[ I=\frac{4.91\times 10^{-6}} {3.14\times 10^{-3}} \] \[ I=1.56\times 10^{-3}\,\text{A} \] \[ I=1.56\,\text{mA} \]
Step 6: Final conclusion.
Therefore, the current in the solenoid is \[ \boxed{1.56\,\text{mA}} \]
Since the charged particle moves in a circular path inside the magnetic field, the magnetic force provides the centripetal force.
Therefore, \[ qvB=\frac{mv^2}{r} \] So, \[ B=\frac{mv}{qr} \]
Step 2: Convert the given quantities into SI units.
Given, \[ m=2.2\times 10^{-30}\,\text{kg} \] \[ q=1.6\times 10^{-19}\,\text{C} \] \[ v=10\,\text{km s}^{-1}=10\times 10^3=10^4\,\text{m s}^{-1} \] \[ r=2.8\,\text{cm}=2.8\times 10^{-2}\,\text{m} \]
Step 3: Calculate the magnetic field.
\[ B=\frac{(2.2\times 10^{-30})(10^4)} {(1.6\times 10^{-19})(2.8\times 10^{-2})} \] \[ B=\frac{2.2\times 10^{-26}} {4.48\times 10^{-21}} \] \[ B=4.91\times 10^{-6}\,\text{T} \]
Step 4: Use magnetic field inside a solenoid.
For a long solenoid, \[ B=\mu_0 n I \] So, \[ I=\frac{B}{\mu_0 n} \] The number of turns per unit length is \[ n=25\,\text{turns/cm} \] Since, \[ 1\,\text{cm}=10^{-2}\,\text{m} \] we get \[ n=25\times 100=2500\,\text{turns/m} \]
Step 5: Calculate the current.
\[ I=\frac{4.91\times 10^{-6}} {(4\pi\times 10^{-7})(2500)} \] Using \[ \pi \approx 3.14 \] \[ I=\frac{4.91\times 10^{-6}} {3.14\times 10^{-3}} \] \[ I=1.56\times 10^{-3}\,\text{A} \] \[ I=1.56\,\text{mA} \]
Step 6: Final conclusion.
Therefore, the current in the solenoid is \[ \boxed{1.56\,\text{mA}} \]
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