Answer:
Explanation:
Given that,
Radius r = 15cm = 0.15m
Area of the circular loop can be determined using the formula for area of a circle
A = π r²
A = π × 0.15²
A = 0.0708 m²
Magnetic field B = 1.2T in positive z direction
B = 1.2 •k T.
If loop is remove from the field in the time interval
∆t = 2.3ms = 2.3×10^-3s
We want to find the average EMF and it is given as
ε = —∆Φ/∆t
The final flux is zero
Φf = 0
Where magnetic flux is given as
Φi = BA Cosθ
Where θ=0 since the area and the magnetic field point in the same direction.
Φi = BA Cos0
Φi = BA
Φi = 1.2 × 0.0708
Φi = 0.0848 Vs
Then, ε = —∆Φ/∆t
ε = —(Φf — Φi) / ∆t
ε = —(0-0.0848) / (2.3×10^-3)
ε = 0.0848 / (2.3×10^-3)
ε = 36.88 V
The EMF is 36.88 Volts
In order to find the our own velocity with respect to land,we need to apply the theory of relative velocity.
Now consider the velocity of the ship traveling towards the north with respect to land as A.Consider our own velocity headed northwards as B.
The relative velocity is the velocity that the body A would appear to an observer on the body B and vice versa.
In this case the relative velocity would be arrived by summing up our velocity with the velocity of the ship as the object (I) is travelling in the ship.
Relative velocity = Velocity of Body A+ Velocity of Body B.
Velocity of the ship traveling towards the north with respect to land(A)= 13.0m/s. (Given)
Our own velocity headed northwards(B)= 2.8 m/s.
Relative velocity = Velocity of Body A+ Velocity of Body B.
Relative velocity= 13.0 + 2.8 = 15.8m/s.
Thus our own velocity with respect to the land is 15.8 m/s.
Answer:
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon)
A photon is a quantum of EM radiation. Its energy is given by E = hf and is related to the frequency f and wavelength λ of the radiation by. E=hf=hcλ(energy of a photon) E = h f = h c λ (energy of a photon) , where E is the energy of a single photon and c is the speed of light.
To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.
The drift velocity is given by the equation:
Where
I = current
n = Number of free electrons
A = Cross-Section Area
q = charge of proton
Our values are given by,
The hall voltage is given by
Where
B= Magnetic field
n = number of free electrons
d = distance
e = charge of electron
Then using the formula and replacing,
Answer:
108.7 V
Explanation:
Two forces are acting on the particle:
- The external force, whose work is 
- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: 
where
q is the charge
is the potential difference
The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:
and since the charge starts from rest, 
, so the formula becomes
In this problem, we have
is the work done by the external force
is the charge
is the final kinetic energy
Solving the formula for , we find
