Answer:
Part a)

Part b)

Explanation:
As we know that magnetic flux through the loop is given as

now we have

now rate of change in flux is given as

now we know that



Now plug in all data


Part b)
Now the radius of the loop after t = 1 s



Now plug in data in above equation


Answer:
c. dioptre that's the answer.
Answer:
Explanation:
Charge on uranium ion = charge of a single electron
= 1.6 x 10⁻¹⁹ C
charge on doubly ionised iron atom = charge of 2 electron
= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C
Let the required distance from uranium ion be d .
force on electron at distance d from uranium ion
= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²
force on electron at distance 61.10 x 10⁻⁹ - r from iron ion
= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²
For equilibrium ,
9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²
2 d² = (61.10 x 10⁻⁹ - r )²
1.414 r = 61.10 x 10⁻⁹ - r
2.414 r = 61.10 x 10⁻⁹
r = 25.31 nm .
Answer:
Explanation:
Let electric potential at A ,B and C be Va , Vb and Vc respectively.
Work done = charge x potential difference
Wab = q ( Va - Vb )
Wac = q ( Va - Vc )
Given
Wac = - Wab / 3
3Wac = - Wab
Now
Wbc = q ( Vb - Vc )
= q [ ( Va-Vc ) - ( Va - Vb )]
= Wac - Wab
= Wac + 3Wac
= 4Wac
Work can be defined as the energy transferred from a body to its sorroundings, the energy spent to move a body, or the energy you need to alter a charged particle, so no energy, no work; thus, the statement is true.