A proton with a speed of $4.0 \times 1 0^{6} m / s$ enters a uniform magnetic field of $0.4 T$ at an angle of $6 0^{c} i r c$ to the field direction. What is the radius of the circular component of its helical trajectory?

hard 5 marks Class 12 · Physics · Moving Charges and Magnetism

Answer

9.1 × 10⁻² m

The correct answer is 9.1 × 10⁻² m — Rounding to two significant figures gives $r \approx 9.1 \times 1 0^{- 2} m$ .

Step-by-step solution

Only the component of velocity perpendicular to the magnetic field produces circular motion. The perpendicular component is $v_{p er p} = v sin θ$ .
Compute $v_{p er p} = 4.0 \times 1 0^{6} m / s \times sin 6 0^{c} i r c = 4.0 \times 1 0^{6} \times 0.866 = 3.464 \times 1 0^{6} m / s$ .
The radius of circular motion for a charged particle is $r = \frac{m v _{p er p}}{q B}$ .
For a proton $m_{p} = 1.67 \times 1 0^{- 27} k g$ and $q = 1.60 \times 1 0^{- 19} C$ .
Substitute: $r = \frac{1.67 \times 1 0 ^{- 27} k g \times 3.464 \times 1 0 ^{6} m / s}{1.60 \times 1 0 ^{- 19} C \times 0.4 T}$ .
Compute numerator: $1.67 \times 1 0^{- 27} \times 3.464 \times 1 0^{6} \approx 5.79 \times 1 0^{- 21} k g m / s$ .
Denominator: $1.60 \times 1 0^{- 19} \times 0.4 = 6.4 \times 1 0^{- 20} C T$ .
Divide: $r = 5.79 \times 1 0^{- 21} /6.4 \times 1 0^{- 20} \approx 0.0905 m$ .
Rounding to two significant figures gives $r \approx 9.1 \times 1 0^{- 2} m$ .

Common mistake: Using the total speed instead of the perpendicular component

v sin θ

in the radius formula

Concept tested: Helical motion of a charged particle in a magnetic field