Answer:
Explanation:
When a wire of length L is moved with velocity v , perpendicular to magnetic field B , EMF will be produced at its two ends . EMF can be calculated with the help of following expression
EMF = BLv
As per this formula , EMF produced will not depend upon number of electrons in the magnetic field.
So B is the right choice.
<h2>The K.E of the charge is 1.02 x 10⁻¹⁷ J</h2>
Explanation:
When the charge of 2e is placed in between the plates .
The force applied on this charge by plates is = q E
here q is the magnitude of charge = 2 e = 2 x 1.6 x 10⁻¹⁹ C
and E is the magnitude of electric field intensity
The work done = Force x displacement
Thus W = q E x S
here S is displacement
Therefore W = 2 x 1.6 x 10⁻¹⁹ x 4 x 8
= 1.02 x 10⁻¹⁷ J
This work will be converted into the kinetic energy of charge .
Thus K.E = 1.02 x 10⁻¹⁷ J
When a wire is moved inside uniform magnetic field then its free electrons will experience magnetic force on it due to which wire will have potential difference at its ends.
Now here we will have magnetic field due to earth and wire is moving in this constant field so induced emf is given by formula
given that
now by using the above formula we will have
Gravitational potential energy=mass*gravitational acceleration*height
Kinetic energy = 0.5*mass*velocity²
So with the given data
K.E 0.5*1*x²=12.5 v²=12.5÷(0.5*1)
v=√12.5÷(0.5*1) v=5
GPE mass*gravitational acceleration*height
1*9.81*h=98
h=98÷(9.81*1)
h= 9.98 m
1) G
2) E
3) D
4) I
5) J
6) C
7) H
8) F
9) B
10) A
I think... i am not 100% sure....
