Answer:
I2>I1
Explanation:
This problem can be solved by using the parallel axis theorem. If the axis of rotation of a rigid body (with moment of inertia I1 at its center of mass) is changed, then, the new moment of inertia is gven by:
where M is the mass of the object and d is the distance of the new axis to the axis of the center of mass.
It is clear that I2 is greater than I1 by the contribution of the term Md^2.
I2>I1
hope this helps!!
Answer:
Inertia = angular momentum / angular velocity
The increasing order of the wavelength from left to right are as follows Gamma Rays < X-rays < UV rays < visible rays < infrared rays < microwaves < radio waves.
The electromagnetic spectrum consists of many waves which are made up of electric field and the magnetic field.
It is said that the electromagnetic wave behaves like particle as well as the wave.
The particle like pieces of the wave are called photons which contains the energy of the wave and the electromagnetic waves are primarily decided by the wavelength and the frequency of the wave.
The group of electromagnetic waves are known as electromagnetic spectrum.
The electromagnetic wave in the order of increasing wavelength are Gamma Rays < X-rays < UV rays < visible rays < infrared rays < microwaves < radio waves.
To know more about electromagnetic spectrum, visit,
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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.
