Answer:
Explanation:
Given that,
Radius of solenoid R = 4cm = 0.04m
Turn per length is N/l = 800 turns/m
The rate at which current is increasing di/dt = 3 A/s
Induced electric field?
At r = 2.2cm=0.022m
µo = 4π × 10^-7 Wb/A•m
The magnetic field inside a solenoid is give as
B = µo•N•I
The value of electric field (E) can
only be a function of the distance r from the solenoid’s axis and it give as,
From gauss law
∮E•dA =qenc/εo
We can find the tangential component of the electric field from Faraday’s law
∮E•dl = −dΦB/dt
We choose the path to be a circle of radius r centered on the cylinder axis. Because all the requested radii are inside the solenoid, the flux-area is the entire πr² area within the loop.
E∮dl = −d/dt •(πr²B)
2πrE = −πr²dB/dt
2πrE = −πr² d/dt(µo•N•I)
2πrE = −πr² × µo•N•dI/dt
Divide both sides by 2πr
E =- ½ r•µo•N•dI/dt
Now, substituting the given data
E = -½ × 0.022 × 4π ×10^-7 × 800 × 3
E = —3.32 × 10^-5 V/m
E = —33.2 µV/m
The magnitude of the electric field at a point 2.2 cm from the solenoid axis is 33.2 µV/m
where the negative sign denotes counter-clockwise electric field when looking along the direction of the solenoid’s magnetic field.
Answer:
Joint application development
Explanation:
Joint application development is a computer based system used for development process. It interface between the users and designer of the system in development.
To see which way atmospheric conditions and meteorological phenomena are moving, and how fast.
Also to see whether they were correct yesterday.
Answer:
Part a)
Part b)
Part c)
Part d)
Explanation:
Part a)
While bucket is falling downwards we have force equation of the bucket given as
for uniform cylinder we will have
so we have
now we have
now we have
Part b)
speed of the bucket can be found using kinematics
so we have
Part c)
now in order to find the time of fall we can use another equation
Part d)
as we know that cylinder is at rest and not moving downwards
so here we can use force balance
