Question

A 144-ω light bulb is connected to a conducting wire that is wrapped into the shape of a square with side length of 83.0 cm. This square loop is rotated within a uniform magnetic field of 454 mt. What is the change in magnetic flux through the loop when it rotates from a position where its area vector makes an angle of 30° with the field to a position where the area vector is parallel to the field?

Answer 1

Answer:

0.04 Wb

Explanation:

The flux through the coil at a generic time t is given by

$\Phi = BA cos \theta$

where

B = 454 mT = 0.454 T is the magnetic field intensity

A is the area of the coil

$\theta$ is the angle between the direction of the field B and the normal to the surface of the coil

The side length of the coil is

L = 83.0 cm = 0.83 m

so its area is

$A=L^2=(0.83 m)^2=0.69 m^2$

At the initial instant, the angle is

$\theta=30^{\circ}$

so the magnetic flux through the coil is

$\Phi_i = BAcos \theta_i = (0.454 T)(0.69 m^2)(cos 30^{\circ})=0.27 Wb$

At the final instant, the angle is

$\theta=0^{\circ}$

so the magnetic flux through the coil is

$\Phi_f = BAcos \theta_i = (0.454 T)(0.69 m^2)(cos 0^{\circ})=0.31 Wb$

So, the change in magnetic flux is

$\Delta \Phi = \Phi_f - \Phi_i = 0.31 Wb-0.27 Wb=0.04 Wb$