Answer:
The magnitude of the magnetic torque on the coil is 1.98 A.m²
Explanation:
Magnitude of magnetic torque in a flat circular coil is given as;
τ = NIASinθ
where;
N is the number of turns of the coil
I is the current in the coil
A is the area of the coil
θ is the angle of inclination of the coil and magnetic field
Given'
Number of turns, N = 200
Current, I = 7.0 A
Angle of inclination, θ = 30°
Diameter, d = 6 cm = 0.06 m
A = πd²/4 = π(0.06)²/4 = 0.002828 m²
τ = NIASinθ
τ = 200 x 7 x 0.002828 x Sin30
τ = 1.98 A.m²
Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²
Answer:
The expresion for the flux through the disk is:
Ф = E·πR^2·cos(Θ).
Explanation:
Let's sat the electric field has direction e and the normal to the disk has direction n (bold means vector quantities). So we have:
E=E·e (where E is the magnitud of the electric flied)
A=A·n
The flux for an uniform electric field and a flat surface is:
Ф=E×A
⇒ Ф = E·A·e×n = E·A·cos(angle(e,n)) = E·A·cos(Θ)
Since in this case the area is for a disk of radius R, 
So, Ф = E·πR^2·cos(Θ)
D.) It is unlikely that a specific cause can be determined, but the treatment would likely be the same in either case.
