Answer:
The minimum frequency of the coil is 7.1 Hz
Explanation:
Given;
number of turns, N = 200 turns
cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
magnitude of magnetic field strength, B = 30 x 10⁻³ T
maximum value of the induced emf, E = 8 V
Maximum induced emf is given as;
E = NBAω
where
ω is angular velocity (ω = 2πf)
E = NBA2πf
where;
f is the minimum frequency, measured in hertz (Hz)
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz
Answer:
a) the magnitude of the force is
F= Q(
) and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E = 
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) =
, where p = q × s
E(r) =
where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q(
)
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²