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 1310 turns per meter and a diameter of 5.55 cm , and carries a current of 2.55 A .

Find the magnetic flux through the loop when L= 2.95 cm .

Answer 1

magnetic field of solenoid is given by,

B = _$\mu$0 n I = _$\mu$0 (N/L) I

B = (4$\pi$ x 10^-7 mT/A) [(1310 m^-1)] (2.55 A)

B = 41956.6 x 10^-7 T

B = 4.19 x 10^-3 T

The magnetic flux through the loop when L= 2.95 cm which will be given as :

using a formula, we have

$\Phi$ = B A

where, A = area of the square loop = L²

$\Phi$ = (4.19 x 10^-3 T) [(2.95 x 10^-2 m)²]

$\Phi$ = (4.19 x 10^-3 T) (8.7025 x 10^-4 m²)

$\Phi$ = 36.4 x 10^-7 Wb

$\Phi$ = 3.64 x 10^-6 Wb