Question Detail
Question
A proton has a 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 × 104 A. Find the maximum torque on a proton in a 2.50 T field.
Correct Answer
Option B is the correct answer.
Detailed Explanation
The magnetic dipole moment (μ) of a current loop is given by μ = IA, where I is the current and A is the area of the loop. For a circular loop, A = πr2.Given:Current (I) = 1.05 × 104 ARadius (r) = 0.650 × 10-15 mMagnetic field (B) = 2.50 TFirst, calculate the magnetic dipole moment μ:μ = I * πr2μ = (1.05 × 104 A) * π * (0.650 × 10-15 m)2μ = 1.05 × 104 * π * (0.4225 × 10-30) Am2μ ≈ 1.396 × 10-26 Am2The maximum torque (τmax) on a magnetic dipole in a magnetic field is given by τmax = μB (when the magnetic moment is perpendicular to the magnetic field).τmax = (1.396 × 10-26 Am2) * (2.50 T)τmax ≈ 3.49 × 10-26 NmThe closest option is 3.48 × 10-26 Nm.
Hint
Calculate the magnetic dipole moment of the current loop first, then use the formula for maximum torque on a magnetic dipole in an external magnetic field.