I need help with number 3 on this sheet. Using Ampere\'s Law, find the strength

Question

I need help with number 3 on this sheet.

Using Ampere's Law, find the strength and direction of the magnetic field of a long of straight wire of radius 40.0 cm carrying current 20.0 A at a.0 cm b. 20.0 cm c. 40.0 cm d.60.0 cm e. Draw the graphical representation of the magnetic field strength vs. distance using B, R, and r. 6 parallel copper wires of small but finite radii lie in the same plane. The wires are separated by equal distance d and they carry equal currents in the same direction. a. What is the direction of the magnetic field? b. Find the magnetic field at point P due to all 6 wires A toroid of 45 turns of wire has an inside radius of 10.0 mm and outside radius of 30.0 mm and carries a current of 20.0 A. Find the magnetic field strength of the toroid at following distances. Provide an explanation for any region that does not provide a magnetic field. a. 5.00 mm b. 20.0 mm c. 40.0 mm

Explanation / Answer

n=45,i=20,a=0.010,b=0.030,r=a+b /2 =0.02

Since

B=mu0*n*I/2*pi*r =1.25*10^-6 *45 *20 /2 * 3.14*0.02= 0.009T

(a)5mm from centr is inside , magnetic field inside the toroid can be found by applying Ampere's Law in the regions outside the wire loops of the toroid. After applying Ampere's Law, we see that the magnetic field at any point inside the toroid is zero because the net electric current is zero.

(B) applying Ampere's Law, we see that the magnetic field at any point inside the toroid is not zero because the net electric current is maximum, not zero.

20 mm is between both radius, and as calculated above, B=0.009T

(C) 40mm is outside, the magnetic field outside the toroid can be found by applying Ampere's Law to determine the magnitude of magnetic field in the regions outside the wire loops of the toroid. After applying Ampere's Law, we see that the magnetic field at any point outside the toroid is zero because the net electric current is zero.

Magnetic field outside the toroid is negligible or zero.

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