Question

a charge of 2 *10^-2 coulomb moves at 30 Revolution per second in circle of diameter 80 cm. what is the current link with the circuit?
explanation is must along with ans.

Answer 1

The current link with the circuit is 0.6 A

Explanation:

The quotient of change in charge divided change in time give the value of the current that would be present in the loop or would be linked with the circuit.

Now for the given question the change charge is is given but the value of change of time has to be found out and this can be done using the inverse of the Number of revolutions per second given.

The change of charge is given by $2 \times 10^-2.$

And time would be $\frac{1}{30}$.

so the change of charge divided by change of time will give us

$2 \times 10^{\frac{-2}{\frac{1}{30}}}$

⇒ $2 \times 30 \times 10^-2$

= 0.6 A.

To know more:

A closed coil having 100 turns is rotated in a uniform magnetic field B=4.0 x10-4 T about a diameter which is perpendicular to the field. the angular velocity of rotation is 300 revolutions/min.The area of the coil is 25 cm2 and its resistance is 4.0 Ω. Find (a) the average emf devoloped in half a turn from a position where the coil is perpendicular to the magnetic field,(b) the average emf in a full turn and (c) the net charge displaced in part (a)

https://brainly.in/question/64671

Answer 2

The change of charge is given by 2 \times 10^-2.2×10

−

2.

And time would be \frac{1}{30}

30

1

.

so the change of charge divided by change of time will give us

2 \times 10^{\frac{-2}{\frac{1}{30}}}2×10

30

1

−2

⇒ 2 \times 30 \times 10^-22×30×10

−

2