Why can we write momentum as √(2mK)?

Since K = 5keV, the ion's speed is far below the speed of light (v ll c), so classical kinetic energy K = (p^2)/2m is strictly valid, giving p = √(2mK).

JEE MainPhysics

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+4 marks−1 if incorrectNumericalPrevious-year question

Radius of a Charged Particle's Path in a Magnetic Field

A singly ionized magnesium atom (A24) ion is accelerated to kinetic energy 5 keV and is projected perpendicularly into a magnetic field B of the magnitude 0.5 T. The radius of path formed will be cm.

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Correct answer

The radius of the circular path formed by the magnesium ion in the magnetic field is $10 c m$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

A singly ionized magnesium ion has mass number $A = 24$ , so $m \approx 24 \times 1.66 \times 10^{- 27} kg$ . Charge $q = e \approx 1.6 \times 10^{- 19} C$ . Kinetic energy $K = 5 keV = 5 \times 10^{3} \times 1.6 \times 10^{- 19} J = 8 \times 10^{- 16} J$ . Magnetic field $B = 0.5 T$ . Perpendicular entry means angle $θ = 90^{\circ}$ .

02Find

Radius of circular path $R$ in centimeters (cm).

03Visualise

The charged ion enters a perpendicular magnetic field. The magnetic Lorentz force provides the necessary centripetal force ( $q v B = m v^{2} / R$ ), making the particle trace a circular trajectory in the plane perpendicular to the field lines.

04Strategise

Relate radius to momentum: $R = \frac{p}{q B} = \frac{\sqrt{2 m K}}{q B}$ . Express everything cleanly using $m = 24 m_{p} \approx 24 \times 1.66 \times 10^{- 27} kg$ and $K = 5000 e J$ . Notice the charge $e$ cancels neatly: $R = \frac{\sqrt{2 \cdot (24 \times 1.66 \times 10^{- 27}) \cdot (5000 \times 1.6 \times 10^{- 19})}}{1.6 \times 10^{- 19} \times 0.5}$ . Alternatively using $1 amu \times 1 eV \approx 1.66 \times 10^{- 27} \times 1.602 \times 10^{- 19} kg \cdot J$ .

05Execute

Evaluate numerator: $2 m K = 2 \times (24 \times 1.66 \times 10^{- 27}) \times (5000 \times 1.6 \times 10^{- 19}) = 48 \times 5000 \times 1.66 \times 1.6 \times 10^{- 46} = 240000 \times 2.656 \times 10^{- 46} = 6.3744 \times 10^{- 41} {kg}^{2} m^{2} / s^{2}$ . Taking the square root gives $p \approx \sqrt{6.4 \times 10^{- 40}} = 8.0 \times 10^{- 21} kg \cdot m/s$ . Denominator: $q B = 1.6 \times 10^{- 19} \times 0.5 = 0.8 \times 10^{- 19} C \cdot T$ . Thus $R = \frac{8.0 \times 10^{- 21}}{0.8 \times 10^{- 19}} = 0.1 m = 10 cm$ .

✓Verify

$R = 0.1 m = 10 cm$ . Units: $[p] / [q B] = (kg \cdot m/s) / (C \cdot T) = m$ . For a low-energy heavy ion in a typical laboratory magnetic field (0.5 T), a path radius of around 10 cm is standard for mass spectrometers.

Hints that build this answer step by step

Which formula correctly relates the radius $r$ of the circular trajectory of a charged particle to its kinetic energy $K$ , mass $m$ , charge $q$ , and magnetic field $B$ ?
$r = \frac{\sqrt{2 m K}}{q B}$
What are the numerical values of the parameters in SI units for a singly ionized magnesium-24 ion ( $A = 24$ )?
$m \approx 24 \times 1.66 \times 10^{- 27} k g$ , $q = 1.6 \times 10^{- 19} C$ , $K = 5 \times 10^{3} \times 1.6 \times 10^{- 19} J$ , $B = 0.5 T$
Substituting the values into $r = \frac{\sqrt{2 m K}}{q B}$ , what is the resulting radius in centimeters?
$10 c m$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

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Students also ask

What is the charge of a singly ionized atom?

A singly ionized atom has lost one electron, so its net charge is $+ 1 e = + 1.6 \times 10^{- 19} C$ .

Why can we write momentum as

\sqrt{2 m K}

Since $K = 5 keV$ , the ion's speed is far below the speed of light ( $v ≪ c$ ), so classical kinetic energy $K = \frac{p^{2}}{2 m}$ is strictly valid, giving $p = \sqrt{2 m K}$ .

Answer

Accepted answer: 10 cm

The radius of the circular path formed by the magnesium ion in the magnetic field is $10 c m$ .

Why each option works or fails

Step-by-step solution

Given: A singly ionized magnesium ion has mass number $A = 24$ , so $m \approx 24 \times 1.66 \times 10^{- 27} kg$ . Charge $q = e \approx 1.6 \times 10^{- 19} C$ . Kinetic energy $K = 5 keV = 5 \times 10^{3} \times 1.6 \times 10^{- 19} J = 8 \times 10^{- 16} J$ . Magnetic field $B = 0.5 T$ . Perpendicular entry means angle $θ = 90^{\circ}$ .
Find: Radius of circular path $R$ in centimeters (cm).
Visualise: The charged ion enters a perpendicular magnetic field. The magnetic Lorentz force provides the necessary centripetal force ( $q v B = m v^{2} / R$ ), making the particle trace a circular trajectory in the plane perpendicular to the field lines.
Strategise: Relate radius to momentum: $R = \frac{p}{q B} = \frac{\sqrt{2 m K}}{q B}$ . Express everything cleanly using $m = 24 m_{p} \approx 24 \times 1.66 \times 10^{- 27} kg$ and $K = 5000 e J$ . Notice the charge $e$ cancels neatly: $R = \frac{\sqrt{2 \cdot (24 \times 1.66 \times 10^{- 27}) \cdot (5000 \times 1.6 \times 10^{- 19})}}{1.6 \times 10^{- 19} \times 0.5}$ . Alternatively using $1 amu \times 1 eV \approx 1.66 \times 10^{- 27} \times 1.602 \times 10^{- 19} kg \cdot J$ .
Execute: Evaluate numerator: $2 m K = 2 \times (24 \times 1.66 \times 10^{- 27}) \times (5000 \times 1.6 \times 10^{- 19}) = 48 \times 5000 \times 1.66 \times 1.6 \times 10^{- 46} = 240000 \times 2.656 \times 10^{- 46} = 6.3744 \times 10^{- 41} {kg}^{2} m^{2} / s^{2}$ . Taking the square root gives $p \approx \sqrt{6.4 \times 10^{- 40}} = 8.0 \times 10^{- 21} kg \cdot m/s$ . Denominator: $q B = 1.6 \times 10^{- 19} \times 0.5 = 0.8 \times 10^{- 19} C \cdot T$ . Thus $R = \frac{8.0 \times 10^{- 21}}{0.8 \times 10^{- 19}} = 0.1 m = 10 cm$ .
Verify: $R = 0.1 m = 10 cm$ . Units: $[p] / [q B] = (kg \cdot m/s) / (C \cdot T) = m$ . For a low-energy heavy ion in a typical laboratory magnetic field (0.5 T), a path radius of around 10 cm is standard for mass spectrometers.

Radius of a Charged Particle's Path in a Magnetic Field

What feels right?

The radius of the circular path formed by the magnesium ion in the magnetic field is $10 c m$ .

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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