A singly ionized magnesium atom (A = 24) 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.

Solution

Physics sir · 4 steps

You are given kinetic energy, not speed, so write mv = √2mK and put it straight into r = mv/qB.

1. Radius from momentum. The magnetic force qvB supplies the centripetal force mv²/r, so r = mv/qB. Since K = p²/2m, the momentum is mv = √2mK.

   r = √2mK/qB

2. SI values. m = 24 × 1.67 × 10⁻²⁷ = 4.008 × 10⁻²⁶ kg. Singly ionized means q = 1.6 × 10⁻¹⁹ C. K = 5 × 10³ × 1.6 × 10⁻¹⁹ = 8 × 10⁻¹⁶ J.
3. Put in numbers. √2mK = √(2 × 4.008 × 10⁻²⁶ × 8 × 10⁻¹⁶) = √(6.41 × 10⁻⁴¹) ≈ 8.0 × 10⁻²¹ kg m s⁻¹. The denominator is qB = 1.6 × 10⁻¹⁹ × 0.5 = 8 × 10⁻²⁰.
4. So. r = (8.0 × 10⁻²¹)/(8 × 10⁻²⁰) ≈ 0.1 m, which is 10 cm.

The common slip: 0.1. Stopping at the SI answer gives 0.1, which is in metres. The question asks for centimetres, so multiply by 100.

The idea to keep. Whenever a charged particle enters a field with a given kinetic energy, use r = √2mK/qB. Convert eV to joules and mass number to kg before you substitute.

Answer from the NTA final answer key. Solution by PrepWiser.