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:
yeah
Explanation:
electric fields help the charged particles interact
and isn't magnetic field the same as electric field.
Answer:
Compose a paragraph that identifies and describes each of the three Siberian regions.
Explanation:
F = m . g = 76.5 x 9..8 = 749.7
Net Force = 3225 - 749.7 = 2475.3
F = m.a
2475.3 = 76.5 a
a = 32.35
V = at + v1
V = at + 0
V = 32.35 x 0.15
V = 4.8525
Hope this helps
