1) The magnitude of the magnetic field at the center of the coil is 0.0609 T. 2) The magnitude of the magnetic field at a point on the axis of the coil a distance of 8.20cm from its center is [tex]7.82 * 10^{-6} T[/tex]
1) The magnetic field at the center of the coil can be calculated using the formula:
[tex]B = \mu_0 * (N * I) / (2 * R)[/tex],
where [tex]\mu_0[/tex] is the permeability of free space [tex](4\pi * 10^{-7} T.m/A)[/tex], N is the number of turns in the coil (410), I is the current flowing through the coil (0.600 A), and R is the radius of the coil (half the diameter, 3.40 cm/2 = 1.70 cm = 0.017 m).
Plugging in these values:
[tex]B = (4\pi * 10^{-7} T.m/A) * (410 * 0.600 A) / (2 * 0.017 m) = 0.0609 T[/tex]
2) For calculating the magnetic field at a point on the axis of the coil, a distance of 8.20 cm from its center, we can use the formula:
[tex]B = \mu_0 * (N * I * R^2) / (2 * (R^2 + d^2)^(3/2))[/tex],
where d is the distance of the point from the center of the coil (8.20 cm = 0.082 m).
Plugging in the values:
[tex]B = (4\pi * 10^{-7} T.m/A) * (410 * 0.600 A * (0.017 m)^2) / (2 * ((0.017 m)^2 + (0.082 m)^2)^(3/2)) = 7.82 * 10^{-6} T[/tex]
Learn more about magnetic fields here:
https://brainly.com/question/30331791
#SPJ11
The complete question is:
A closely wound, circular coil with a diameter of 3.40 cm has 410 turns and carries a current of 0.600A
1) What is the magnitude of the magnetic field at the center of the coil?
2) What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 8.20cm from its center?