Question

A single-turn square loop of side L is centered on the axis of a long solenoid. In addition, the plane of the square loop is perpendicular to the axis of the solenoid. The solenoid has 1210 turns per meter and a diameter of 6.25 cm , and carries a current of 3.00 A .

Find the magnetic flux through the Loop when L= a)2.70 b)6.25 c)14.0

Answer 1

Magnetic field through solenoid is given by:

B = u0*N*i/L

B = 4*pi*10^-7*1210*3/1 = 4.56*10^-3 T

Now magnetic flux is given by:

phi = B*A*cos X

Since X = 0 deg

cos 0 = 1

So,

phi = B*A

1.

I am assuming length L is in cm

then L = 2.7 cm = 0.027 m

A = L^2 = 0.027^2

phi = 4.56*10^-3*(0.027^2) = 3.32*10^-6 Wb

2.

Since magnetic field is contained only within the solid, so

Max magnetic flux possible will be for max square loop possible inside the solenoid

side of the loop will be = sqrt (d/2) = sqrt (6.25^2/2) = 4.416 cm

when L = 4.416 cm = 0.04416 m

then magnetic flux will be

phi = 4.56*10^-3*(0.04416^2) = 8.89*10^-6 Wb

3.

Magnetic flux max will be same as part 2

phi = 8.89*10^-6 Wb