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
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fault. Normal (extensional ) fault is a
displacement of a rock as a result of rock-mass movement and occurs when the
crust is stretching. Because of the stretching the thickness of the crust is
reduced and the crust or horizontally extended. </span>
the above three pictures may help you
go through the attachments
A pendulum is not a wave.
-- A pendulum doesn't have a 'wavelength'.
-- There's no way to define how many of its "waves" pass a point
every second.
-- Whatever you say is the speed of the pendulum, that speed
can only be true at one or two points in the pendulum's swing,
and it's different everywhere else in the swing.
-- The frequency of a pendulum depends only on the length
of the string from which it hangs.
If you take the given information and try to apply wave motion to it:
Wave speed = (wavelength) x (frequency)
Frequency = (speed) / (wavelength) ,
you would end up with
Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz
Have you ever seen anything that could be described as
a pendulum, swinging or even wiggling back and forth
85 times every second ? ! ? That's pretty absurd.
This math is not applicable to the pendulum.