Question

13. A 12-cm-diameter circular loop of wire is placed in a 1.19-T magnetic field.

a.) When the plane of the loop is perpendicular to the field lines, what is the magnetic flux through the loop?

b.) The plane of the loop is rotated until it makes a 35 ∘ angle with the field lines. What is the angle in the equation ΦB = BAcosθ for this situation?

c.) What is the magnetic flux through the loop at this angle?

Answer 1

given

d = 12 cm = 0.12 m
B = 1.19 T

Area of the loop, A = pi*d^2/4

= pi*0.12^2/4

= 0.0113 m^2

a) magnetic flux through the loop = A*B*cos(theta) (here theta is the angle between normal to the loop and B)

= 0.0113*1.19*cos(0)

= 0.0134 T.m^2

b) we know, theta is the angle between normal to the surface and B.

so, theta = 90 - 35

= 55 degrees

c) magnetic flux through the loop = A*B*cos(theta) (here theta is the angle between normal to the loop and B)

= 0.0113*1.19*cos(55)

= 0.00771 T.m^2