What magnetic field B is needed to keep 998-GeV protons revolving in a circle of radius 1.0km? Use the relativistic mass. The proton’s “rest mass” is 0.938 GeV/c². ( 1 GeV = 10⁹ eV.) [Hint: In relativity, mᵣₑₗ v²/r = qvB is still valid in a magnetic field, where mᵣₑₗ = γm.]

Verified step by step guidance

1. First, calculate the Lorentz factor (γ) using the formula γ = \( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \). Since the total energy E is given as 998 GeV and the rest mass energy (m₀c²) is 0.938 GeV, use the relation E = γm₀c² to find γ: \( γ = \frac{E}{m₀c²} = \frac{998 \text{ GeV}}{0.938 \text{ GeV}} \approx 1064 \).
2. Next, calculate the relativistic mass (mᵣₑₗ) using mᵣₑₗ = γm₀. Here, m₀ (rest mass) = 0.938 GeV/c², so mᵣₑₗ = 1064 × 0.938 GeV/c² ≈ 998 GeV/c².
3. Convert the relativistic mass from GeV/c² to kg. Knowing that 1 GeV/c² = 1.783 × 10⁻²⁷ kg, mᵣₑₗ = 998 GeV/c² × 1.783 × 10⁻²⁷ kg/GeV/c² ≈ 1.78 × 10⁻²⁴ kg.
4. Use the centripetal force formula for a charged particle in a magnetic field, which is mᵣₑₗ v²/r = qvB. Here, rearrange to solve for B: B = \( \frac{mᵣₑₗ v²}{rqv} \). Since v can be approximated by c (speed of light) for high-energy protons, and q (charge of proton) = 1.602 × 10⁻¹⁹ C, substitute these values in: B = \( \frac{(1.78 × 10⁻²⁴ kg)(3 × 10⁸ m/s)²}{(1.0 × 10³ m)(1.602 × 10⁻¹⁹ C)(3 × 10⁸ m/s)} \).
5. Simplify and calculate B: B = \( \frac{(1.78 × 10⁻²⁴ kg)(9 × 10¹⁶ m²/s²)}{(1.602 × 10⁻¹⁹ C)(3 × 10¹¹ m²/s)} \) ≈ 0.31 T. Therefore, a magnetic field of approximately 0.31 Tesla is needed to keep the protons revolving in a circle of radius 1.0 km.