Answer:
I = 2172.46 A
Explanation:
Given that,
The length of a solenoid, l = 2.1 m
The inner radius of the solenoid, r = 28 cm = 0.28 m
The number of turns in the wire, N = 1000
The magnetic field in the solenoid, B = 1.3 T
We need to find the current carried by it. We know that, the magnetic field in a solenoid is given by :

Put all the values,

So, it carry current of 2172.46 A.
Answer:
True
Explanation:
A crowbar makes our work easier by multiply effort because it belongs to first class lever.
And first class lever makes work easier by multiplying the effort
Answer : 5m/s
Explanation:the formular for velocity is distance /time or you can say displacement /time. Then it would then be
100/20 =5m/s
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
Acceleration=(change in speed)/(time for the change). 43/0.28 = 153.6 m/s^2.