Question

In a physics laboratory experiment, a coil with 180 turns enclosing an area of 13.5 cm^2 is rotated during the time interval 5.00×10-2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.00×10-5 T.A) What is the total magnitude of the magnetic flux ( Phi_initial) through the coil before it is rotated?B) What is the magnitude of the total magnetic flux Phi_final through the coil after it is rotated?Express your answer numerically, in webers, to at least three significant figures.C) What is the magnitude of the average emf induced in the coil?Express your answer numerically (in volts) to at least three significant figures

Answer 1

A)
`1.22\cdot 10^(-5) Wb`

The total magnetic flux through the coil before it is rotated is given by:

`\Phi_i = NBA cos \theta`

where

N = 180 is the number of turns in the coil

`B=5.00\cdot 10^(-5)T` is the magnetic field intensity

`A=13 cm^2 = 13\cdot 10^(-4) m^2` is the area enclosed by the coil

`\theta=0^(\circ)` is the initial angle between the direction of the normal to the surface of the coil and the magnetic field

Substituting, we find

`\Phi_i = (180)(5.00\cdot 10^(-5) T)(13.5\cdot 10^(-4) m^2)(cos 0^(\circ))=1.22\cdot 10^(-5) Wb`

B) 0

The total magnetic flux through the coil after it is rotated is still given by:

`\Phi_i = NBA cos \theta`

where this time,

`\theta=90^(\circ)` is the final angle between the direction of the normal to the surface of the coil and the magnetic field

since
`cos \theta=cos 90^(\circ) =0`, we now have that the total flux through the coil is zero.

C)
`2.44\cdot 10^(-4) V`

The magnitude of the average emf induced in the coil is given by

`\epsilon=(\Delta \Phi)/(\Delta t)`

where

`\Delta \Phi = \Phi_i =1.22\cdot 10^(-5)Wb` is the change of total flux

`\Delta t=5.00\cdot 10^(-2) s` is the time taken

Susbtituting, we find

`\epsilon=(1.22\cdot 10^(-5) Wb)/(5.00\cdot 10^(-2)s)=2.44\cdot 10^(-4) V`