Given Information:
Length of wire = 132 cm = 1.32 m
Magnetic field = B = 1 T
Current = 2.2 A
Required Information:
(a) Torque = τ = ?
(b) Number of turns = N = ?
Answer:
(a) Torque = 0.305 N.m
(b) Number of turns = 1
Explanation:
(a) The current carrying circular loop of wire will experience a torque given by
τ = NIABsin(θ) eq. 1
Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.
We know that area of circular loop is given by
A = πr²
where radius can be written as
r = L/2πN
So the area becomes
A = π(L/2πN)²
A = πL²/4π²N²
A = L²/4πN²
Substitute A into eq. 1
τ = NI(L²/4πN²)Bsin(θ)
τ = IL²Bsin(θ)/4πN
The maximum toque occurs when θ is 90°
τ = IL²Bsin(90)/4πN
τ = IL²B/4πN
torque will be maximum for N = 1
τ = (2.2*1.32²*1)/4π*1
τ = 0.305 N.m
(b) The required number of turns for maximum torque is
N = IL²B/4πτ
N = 2.2*1.32²*1)/4π*0.305
N = 1 turn
Answer:
η = 58.8%
Explanation:
Work is defined as the force applied by the distance traveled by the body.

where:
W = work [J] (units of joules)
F = force = 294 [N]
d = distance = 5 [m]
![W = 294*5\\W = 1470 [J]\\](https://tex.z-dn.net/?f=W%20%3D%20294%2A5%5C%5CW%20%3D%201470%20%5BJ%5D%5C%5C)
Efficiency is defined as the energy required to perform an activity in relation to the energy actually added to perform some activity. This can be better understood by means of the following equation.

The formula for average speed is S=D/T
1. S=72m/37s
Divide
S= 1.94
Kira's average speed is 1.94m/s.
2. S=7.5km / 1.5h
S=5
Your average speed is 5km/h
3. S=1260km/3.5h
S=360
The airplanes average speed is 360km/h
Answer:
1.876 J
Explanation:
First, let’s calculate the compression of the spring from the Hooke’s law:
F=kx,
here, F=75 N is the force acted on the spring, k=1500 N⁄m is the force constant of the spring, x is the compression of the spring.
Then, we get:
x=F/k=(75 N)/(1500 N/m)=0.05 m.
Finally, we can find the potential energy stored in the spring:
PE=1/2 kx^2=1/2∙1500 N/m∙(0.05 m)^2=1.875 J.
correct my answer if it's wrong ^^