Question

A typical magnitude of the external magnetic field in a cardiac catheter ablation procedure using remote magnetic navigation is B = 0.080 T. Suppose that the permanent magnet in the catheter used in the procedure is inside the left atrium of the heart and subject to this external magnetic field. The permanent magnet has a magnetic moment of 0.17 A · m2. The orientation of the permanent magnet is 27° from the direction of the external magnetic field lines.

(a) What is the magnitude of the torque on the tip of the catheter containing this permanent magnet?
(b) What is the potential energy of the system consisting of the permanent magnet in the catheter and the magnetic field provided by the external magnets?

Answer 1

Step-by-step explanation:

B, magnitude of the external magnetic field = 0.080 T

μ, magnitude of magnetic moment = 0.17 Am²

θ, angle between the magnetic moment vector and the external magnetic field = 27°

τ, magnitude of the torque = μB sinθ = 0.17 Am² × 0.08 × sin 27° = 0.00617 Nm

potential energy = -μBcosθ = -0.17 A m² × 0.08 × cos 27° = 0.0121 J