Answer:
a. φ = 0.867 x 10⁻³ W
b. φ = 9.542 x 10⁻³ W
c. I = 41.6 A
d. Direction oppose the increase field
Explanation:
d = 47 cm , R = 120 Ω , β = 5.0 mT t = 0s , β = 55 mT t = 25 ms
a.
φ = β * A ⇒ φ = β * ¹/₄ * π * d²
φ = 5.0 mT * ¹/₄ * π * (470 x 10 ⁻³m )² = 0.867 x 10⁻³ W
b.
φ = β * A ⇒ φ = β * ¹/₄ * π * d²
φ = 55 mT * ¹/₄ * π * (470 x 10 ⁻³m )² = 9.542 x 10⁻³ W
c.
Δt = (9.541 - 0.867) x 10⁻³ = 8.675 x 10⁻³ W / 25 ms
E = 0.347 V
I = E * R ⇒ I = 0.347v * 120 Ω = 41.6 A
d.
The direction must oppose the increase of the external field downward
This question involves the concepts of magnetic flux, Ohm's Law, induced current, and right-hand rule.
A) The magnetic flux through the loop at the beginning is "0.867 x 10⁻³ web".
B) The magnetic flux through the loop at the end is "9.52 x 10⁻³ web".
C) The loop's induced current during this time is "2.88 mA"
D) The direction of flow of current is "clockwise".
A)
The magnetic flux at the beginning of the loop is given by the following formula:
[tex]\phi = BA\ Cos\ \theta[/tex]
where,
[tex]\phi[/tex] = magnetic flux at the beginning of the loop = ?
B = Magnetic field at the beginning of the loop = 5 mT = 0.005 T
A = Cross-sectional area =[tex]\pi\frac{d^2}{4}=\pi \frac{(0.47\ m)^2}{4} =[/tex] 0.173 m²
θ = angle between the normal to area and the magnetic field = 0°
Therefore,
[tex]\phi=(0.005\ T)(0.173\ m^2)Cos0^o\\\phi = 0.867\ x\ 10^{-3}\ Web[/tex]
B)
The magnetic flux at the end of the loop is given by the following formula:
[tex]\phi = BA\ Cos\ \theta[/tex]
where,
[tex]\phi[/tex] = magnetic flux at the end of the loop = ?
B = Magnetic field at the end of loop = 55 mT = 0.055 T
A = Cross-sectional area =[tex]\pi\frac{d^2}{4}=\pi \frac{(0.47\ m)^2}{4} =[/tex] 0.173 m²
θ = angle between the normal to area and the magnetic field = 0°
Therefore,
[tex]\phi=(0.055\ T)(0.173\ m^2)Cos0^o\\\phi = 9.52\ x\ 10^{-3}\ Web[/tex]
C)
First, we will find out the induced E.M.F using the following formula:
[tex]V =\frac{\Delta\phi}{\Delta t}\\\\V = \frac{(9.52 - 0.867)\ x\ 10^{-3}\ web}{25\ ms}\\\\V = \frac{(9.52 - 0.867)\ x\ 10^{-3}\ web}{25\ x\ 10^{-3} s}[/tex]
V = 0.35 volt
Using Ohm's Law to find out induced current:
[tex]V=IR\\\\I=\frac{V}{R}[/tex]
where,
I = induced current = ?
R = Resistance = 120 Ω
Therefore,
[tex]I=\frac{0.35\ volt}{120\ \Omega}[/tex]
I = 2.88 x 10⁻³ A = 2.88 mA
D)
The right-hand rule states that if the thumb of the right hand points in the direction of the magnetic field, then the direction of the curl of fingers of the right hand gives the direction of the current in coils.
Applying the right-hand rule to this scenario gives the direction of the induced current as clockwise.
Learn more about magnetic flux here:
https://brainly.com/question/24615998?referrer=searchResults
The attached picture shows the magnetic flux.
