Answer:
The magnitude of the field is 8.384×10^-4 T.
Explanation:
Now, i start solving this question:
First, convert the potential difference(V) 2 kv to 2000 v.
As, we have the final formula is qvB = mv^2/r. It came from the centripetal force and the magnetic force and we know that these two forces are equal. When dealing with centripetal motion use the radius and not the diameter so
r = 0.36/2 = 0.18 m.
As, we are dealing with an electron so we know its mass is 9.11*10^-31 kg and its charge (q) is 1.6*10^-18 C.
We can solve for its electric potential energy by using ΔU = qV and we know potential energy initial is equal to kinetic energy final so ΔU = ΔKE and kinetic energy is equal to 1/2mv^2 J.
qV = 1/2mv^2
(1.6*10^-19C)(2000V) = (1/2)(9.11*10^-31kg) v^2
v = 2.65×10^7 m/s.
These all above steps we have done only for velocity(v) because in the final formula we have 'v' in it. So, now we substitute the all values in that formula and will find out the magnitude of the field:
qvB = mv^2/r
qB = mv/r
B = mv/qr
B = (9.11*10^-31 kg)(2.65×10^7 m/s) / (1.6*10^-19 C)(0.18 m)
Hence, B = 8.384*10^-4 T.
Answer:
Elastic potential energy is stored when materials stretch or compress. Examples of it include springs, rubber bands, and slingshots.
Explanation:
I think the answer for the question above its b 1.2
Answer:
Explanation:
Number of turns
N = 210turns
Length of solenoid
l = 0.18m
Cross sectional area
A = 4cm² = 4 × 10^-4m²
A. Inductance L?
Inductance can be determined using
L = N²μA/l
Where
μ is a constant of permeability of the core
μ = 4π × 10^-7 Tm/A
A is cross sectional area
l is length of coil
L is inductance
Therefore
L = N²μA / l
L=210² × 4π × 10^-7 × 4 × 10^-4 / 0.18
L = 1.23 × 10^-4 H
L = 0.123 mH
B. Self induce EMF ε?
EMF is given as
ε = -Ldi/dt
Since rate of decrease of current is 120 A/s
Then, di/dt = —120A/s, since the current is decreasing
Then,
ε = -Ldi/dt
ε = - 1.23 × 10^-4 × -120
ε = 0.01478 V
ε ≈ 0.015 V
Answer:
The electric field outside the sphere will be 
.
Explanation:
Given that,
Radius of solid sphere = R
Charge = q
According to figure,
Suppose r is the distance between the point P and center of sphere.
If 
be the volume charge density,
Then, the charge will be,
.....(I)
Consider a Gaussian surface of radius r.
We need to calculate the electric field outside the sphere
Using formula of electric field
Put the value from equation (I)
Hence, The electric field outside the sphere will be .
