Please show your work. 1.) Two circular coils of current-carrying wire have the
ID: 1394702 • Letter: P
Question
Please show your work.
1.) Two circular coils of current-carrying wire have the same magnetic moment. The first coil has a radius of 0.097 m, has 139 turns, and carries a current of 4.7 A. The second coil has 158 turns and carries a current of 9.1 A. What is the radius of the second coil?
_______m
2.) Two pieces of the same wire have the same length. From one piece, a square coil containing a single loop is made. From the other, a circular coil containing a single loop is made. The coils carry different currents. When placed in the same magnetic field with the same orientation, they experience the same torque. What is the ratio Isquare / Icircle of the current in the square coil to the current in the circular coil?
Explanation / Answer
1) Formula For magnetic moment is mu = N*I*A
N is the number of turns
I is the current in the coil
A is the cross sectional area of the coil = pi*r^2
r is the radius of the coil
since the magnetic moments of the two coils are same then
N1*I1*pi*r1^2 = N2*I2*pi*r2^2
139*4.7*3.142*0.097^2 = 158*9.1*3.142*r2^2
19.3 = 4517.5*r2^2
r2^2 = 19.3/4517.5
r2 = sqrt(0.004272) =0.0653 m
2) Torque T = mu*B*sin(theta)
B is the magnetic field
theta is the angle of orientataion
mu is the magnetic moment = N*I*A
N is the no.of turns = 1
I is the current in the coil
A is the area of cross section
let L be the length of the wire
then length of the each side of the square = L/4
area of the square = (L/4)*(L/4) = L^2/16
circumference of the circle = 2*pi*r = L
radius of the coil is r = L/(2*pi).....
Area of the coil is A = pi*r^2 = pi*(L^2)/(4*pi^2) = L^2/(4*pi)
Given torque and magnetic field both are same then magnetic moment is also same
Isquare*Asquare = Icircle*Acircle
Isquare/Icircle = Acircle/Asquare = [L^2/(4*pi)]/[L^2/16] = 16/(4*pi) = 4/3.142 = 1.27