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 0.14 Am^2. The orientation of the permanent magnet is 22° 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?
To find the magnitude of the torque on the tip of the catheter, you can use the formula:
τ = B * μ * sin(θ)
where τ is the torque, B is the magnitude of the external magnetic field, μ is the magnetic moment of the permanent magnet, and θ is the angle between the magnetic moment and the direction of the external magnetic field lines.
Substituting the given values:
B = 0.080 T
μ = 0.14 A*m^2
θ = 22°
First, convert θ from degrees to radians:
θ_radians = θ * π/180
θ_radians = 22° * π/180
θ_radians = 0.384 radians
Now, plug in the values into the formula:
τ = 0.080 T * 0.14 A*m^2 * sin(0.384 radians)
Calculate the sine value:
sin(0.384 radians) ≈ 0.375
Lastly, calculate the torque:
τ ≈ 0.080 T * 0.14 A*m^2 * 0.375
The torque will be in units of N*m (Newton-meters).
To find the potential energy of the system, you can use the formula:
U = -μ * B * cos(θ)
where U is the potential energy.
Substituting the given values:
B = 0.080 T
μ = 0.14 A*m^2
θ = 22°
First, convert θ from degrees to radians:
θ_radians = θ * π/180
θ_radians ≈ 0.384 radians
Now, plug in the values into the formula:
U = -0.14 A*m^2 * 0.080 T * cos(0.384 radians)
Calculate the cosine value:
cos(0.384 radians) ≈ 0.927
Lastly, calculate the potential energy:
U ≈ -0.14 A*m^2 * 0.080 T * 0.927
The potential energy will be in units of Joules (J).