Explanation:
It is given that, the field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs
Change in magnetic field,
Change in time,
Diameter, d = 2.3 cm
Radius, r = 0.0115 m
Emf is induced in the ring as the field changes. It is given by :
E = 5.19 volts
So, the emf induced in the ring is 5.19 volts. Hence, this is the required solution.
Answer:
Explanation:
Given
Sphere of Radius R
Suppose mass of block is m
At any instant \theta Normal reaction(N) and weight(mg) is acting such that
, where v is velocity of block at any angle \theta
When block is just about to leave then N=0
therefore
-------------------1
Also by conserving Energy we get
Potential Energy=kinetic Energy of block
here h=vertical distance traveled by block
From diagram
-----------------2
From 1 and 2
Thus from this value of h is
The frequency of bird chirping hear by hiran will be 1.77 kHz.
<u>Explanation:</u>
As per Doppler effect, the observer will feel a decrease in the frequency of the receiving signal if the source is moving away from the observer. So the shifted frequency is obtained using the below equation:
Here , c is the speed of sound, Vs is the velocity of source with which it is moving away. f is the original frequency of source and f' is the frequency shift heard by the observer.
As here, f = 1800 Hz, Vs= 6 m/s and c = 343 m/s, then
So, the frequency of bird chirping hear by hiran will be 1.77 kHz.
Answer:
a) E =0, b) E = 1,129 10¹⁰ N / C , c) E = 3.33 10¹⁰ N / C
Explanation:
To solve this exercise we can use Gauss's law
Ф = ∫ E. dA = / ε₀
Where we must define a Gaussian surface that is this case is a sphere; the electric field lines are radial and parallel to the radii of the spheres, so the scalar product is reduced to the algebraic product.
E A = /ε₀
The area of a sphere is
A = 4π r²
E = q_{int} / 4πε₀ r²
k = 1 / 4πε₀
E = k q_{int} / r²
To find the charge inside the surface we can use the concept of density
ρ = q_{int} / V ’
q_{int} = ρ V ’
V ’= 4/3 π r’³
Where V ’is the volume of the sphere inside the Gaussian surface
Let's apply this expression to our problem
a) The electric field in center r = 0
Since there is no charge inside, the field must be zero
E = 0
b) for the radius of r = 6.0 cm
In this case the charge inside corresponds to the inner sphere
q_{int} = 5.0 4/3 π 0.06³
q_{int} = 4.52 10⁻³ C
E = 8.99 10⁹ 4.52 10⁻³ / 0.06²
E = 1,129 10¹⁰ N / C
c) The electric field for r = 12 cm = 0.12 m
In this case the two spheres have the charge inside the Gaussian surface, for which we must calculate the net charge.
The charge of the inner sphere is q₁ = - 4.52 10⁻³ C
The charge for the outermost sphere is
q₂ = ρ 4/3 π r₂³
q₂ = 8.0 4/3 π 0.12³
q₂ = 5.79 10⁻² C
The net charge is
q_{int} = q₁ + q₂
q_{int} = -4.52 10⁻³ + 5.79 10⁻²
q_{int} = 0.05338 C
The electric field is
E = 8.99 10⁹ 0.05338 / 0.12²
E = 3.33 10¹⁰ N / C