Question

In a physics laboratory experiment, a coil with 250 turns enclosing an area of 11.6 cm2 is rotated during the time interval 3.90×10−2 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 7.00×10−5 T .

Answer 1

Answer:

a) Ф$_{B.i}$ =  $8.1 X 10^{-8} Wb$, Ф$_{B.f}$ = 0

b) E = 0.52mV

Explanation:

a) considering the loop of wire which has an area of A = 11.6cm² and N = 250 turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in Δt = 0.040s. Given that the earths magnetic field at the position of the loop is B = 7.0 x $10^{-5 } T$, the flux through the loop before it is rotated is:

Ф$_{B.i}$ = BAcos(Ф$_{i}$) = BAcos(0°)

Ф$_{B.i}$ =(7.0 x $10^{-5 } T$)(11.6 x $10^{-4} m^{2}$)(1)

Ф$_{B.i}$  = $8.1 X 10^{-8} Wb$

Ф$_{B.i}$ =  $8.1 X 10^{-8} Wb$

after it is rotated the angle between the area and the magnetic field is Ф = 90°, hence

Ф$_{B.f}$ = BAcos(Фf) = BAcos(90°) = 0

b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns that is

|Eav| = N | ФB,f - ФB,i |/ Δt

$=|E_{av} | =250 \frac{0-8.1X 10^{-8} }{3.90 X 10^{-2} }$

= 5.2 x $10^{-4} V$

E = 0.52mV

Answer 2

Answer:

a) Φin = 8.12x10⁻⁸ Wb

b) Φfin = 0

c) ε = 5.2x10⁻⁴ V

Explanation:

a) The total magnitude of magnetic flux before it is rotated is equal

Φin = B*A*sinθ

B = 7x10⁻⁵ T

A = 11.6 cm² = 0.00116 m²

Replacing

Φin = 7x10⁻⁵ * 0.00116 * sin90° = 8.12x10⁻⁸ Wb

b) The toal magnitude of magnetic flux after it is rotated

Φfin = B*A*sinθ = 7x10⁻⁵ * 0.00116 * sin0° = 0

c) Magnitude of the average emf induced is equal to:

$e = N(\frac{O_{in}-O_{fin} }{t} )=250(\frac{8.12x10^{-8}-0 }{3.9x10^{-2} } )=5.2x10^{-4} V$