Answer:
Potential difference and charge will also increase.
Explanation:
Asking that :
What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unchanged?
The charge is directly proportional to area of the plate. That is, increase in area of the plate of a capacitor will lead to the increase in the charges between the plates.
And since charge is also proportional to the magnitude of potential difference between the plates from the definition of capacitance of a capacitor which says that:
Q = CV
Therefore, increase in the area of the plate will also lead to increase in potential difference between the plates.
Therefore, if the plate area were increased while the plate separation remains unchanged, the charge and potential difference between them will also increase.
Answer:
= 591.45 T/s
Explanation:
i = induced current in the loop = 0.367 A
R = Resistance of the loop = 117 Ω
E = Induced voltage
Induced voltage is given as
E = i R
E = (0.367) (117)
E = 42.939 volts
= rate of change of magnetic field
A = area of loop = 7.26 x 10⁻² m²
Induced emf is given as
= 591.45 T/s
Answer:
a mixture that can be separated
Answer:
ΔΦ = -3.39*10^-6
Explanation:
Given:-
- The given magnetic field strength B = 0.50 gauss
- The angle between earth magnetic field and garage floor ∅ = 20 °
- The loop is rotated by 90 degree.
- The radius of the coil r = 19 cm
Find:
calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.
Solution:
- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.
- The strength of magnetic field B and the are of the loop A remains constant. So we have:
Φ = B*A*cos(θ)
ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )
- The initial angle θ_1 between the normal to the coil and B was:
θ_1 = 90° - ∅
θ_1 = 90° - 20° = 70°
The angle θ_2 after rotation between the normal to the coil and B was:
θ_2 = ∅
θ_2 = 20°
- Hence, the change in flux can be calculated:
ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )
ΔΦ = -3.39*10^-6
