Question Detail
Question
A proton has magnetic field due to its spin on its axis. The field is similar to that created by a circular current loop 0.650 × 10⁻¹⁵ m in radius with a current of 1.05 x 10⁴ A. Find the maximum torque on a proton in a 2.50T field.
Correct Answer
Option B is the correct answer.
Detailed Explanation
First, calculate the magnetic dipole moment (μ) of the current loop, which simulates the proton's spin magnetic moment:μ = I * A, where A is the area of the loop (πr²).Given: I = 1.05 × 10⁴ A, r = 0.650 × 10⁻¹⁵ m.Area A = π * (0.650 × 10⁻¹⁵ m)² = π * 0.4225 × 10⁻³⁰ m² ≈ 1.3273 × 10⁻³⁰ m².Magnetic moment μ = (1.05 × 10⁴ A) * (1.3273 × 10⁻³⁰ m²) ≈ 1.3937 × 10⁻²⁶ A·m².Next, calculate the maximum torque (τ_max) on the magnetic dipole in a magnetic field (B):τ_max = μ * BGiven: B = 2.50 T.τ_max = (1.3937 × 10⁻²⁶ A·m²) * (2.50 T) ≈ 3.484 × 10⁻²⁶ N·m.This value is closest to 3.48 × 10⁻²⁶ N.m.
Hint
First, calculate the magnetic dipole moment of the equivalent current loop. Then use the formula for maximum torque on a magnetic dipole in a magnetic field.