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Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

Answer of Question no. 1 is

v = 3.0×106 m/s
B = 4.0×10-2 T
q = -1.60×10-19C
F = ?
F = Bvq
F = ( 4.0×10-2T )(3.0×106 m/s ) (-1.60×10-19C )
=( 4.0×10-2N/a.m )( 3.0×106m/s ) ( -1.60×10-19C)
= -(1.9x10-14 )

ase krna hy ++Noor-e-eman

1)He would find the same field, since the extant field does not depend on the charge he is prying with. The extant field is kq / d², q being the unknown charge.

II. He finds different forces with different signs. F = kqQ / d²
where Q is the charge he is prying with.

 Answer of question no.5 is in detail Answer :   Potential of a certain photocell is = 6.0 V  q = -1.60×10-19C

In our MKS units, the unit of potential is 1 Joule 1 Volt. Another useful unit is

Coulomb  "electron volt" or eV

So the Potential of a certain photocell is = 6.0 V

 One electron - volt = energy gained by moving one electron charge through one Volt (1.6x10-19C)x1V = (1.6xC)x6.0V                              = 9.6x10⁻¹⁹ J

Question No 5

The stopping potential of a certain photocell is 6.0 V. What is the kinetic energy given to the
electrons by the incident light? Give your answer in both joules and electron volts

6.0eV

6.0 x 1.6x10⁻¹⁹ = 9.6x10⁻¹⁹ J

Answer of question no. 4 is

 Answer : Resistors in series increase by simple addition. R total = R1 + R2 + R3 ...  (I) 12-V battery in a circuit with three resistors connected in series  the resistance of one of the resistors increases The series resistance will also increase resistance    (II) There is no change in voltage . Battery voltage will remain same   (III)   Current passes in the resistors will decrease as the resistance increase.

Question# 4

Consider a 12-V battery in a circuit with three resistors connected in series.

I. If the resistance of one of the resistors increases, how will the series resistance change?
II. Will there be any change in the battery voltage?
III. What will happen to the current? Marks = 2+2+2 = 6

Resistors in series increase by simple addition.
R total = R1 + R2 + R3 ...
(1) Adding resistance in series increases total resistance.
(2) Increased resistance has no effect on the resistance
of an ideal battery.
Real batteries have internal resistance which consumes
power and drops Voltage in an inverse relation with load
resistance. Increasing load resistance could increase
battery Voltage but not necessarilly measurably.
(3) Increased resistance reduces current.

Question No 3

The whole electrical technology revolves round the devices i.e. motor, generator, and transformer. Write your answer in a single line that under what relationship these devices came into existence

Solution

Induced magnetic fields, and magnetic lines of force are what make motors, generators and transformers possible.

magnetic lines of force are responsible for working of these devices.

A beam of electrons travels at 3.0×106 m/s through a uniform magnetic field of 4.0×10-2 T at right angles to the field. How strong is the force acting on each electron? Marks = 6
(Value of charge on electron is -1.60×10-19C)

This is a plug n chug problem.
The magnetic force on a charged particle is:
F = (q*v) x B
q = charge of particle, v = velocity, B is the mag field
Since they are at a right angle, the equation simplifies to

F = q*v*B

You have these three values and the force will be in Newtons.

Answer of Question no. 1 is

v = 3.0×106 m/s
B = 4.0×10-2 T
q = -1.60×10-19C
F = ?
F = Bvq
F = ( 4.0×10-2T )(3.0×106 m/s ) (-1.60×10-19C )
=( 4.0×10-2N/a.m )( 3.0×106m/s ) ( -1.60×10-19C)
= -(1.9x10-14 )

Question No 01

v = 3.0×106 m/s
B = 4.0×10-2 T
q = -1.60×10-19C
F = ?
F = Bvq
F = ( 4.0×10-2T )(3.0×106 m/s ) (-1.60×10-19C )
=( 4.0×10-2N/a.m )( 3.0×106m/s ) ( -1.60×10-19C)
= -(1.9x10-14 )

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