Question

shown in the following figure is a long, straight wire and a single-turn rectangular loop, both of which lie in the plane of the page. the wire is parallel to the long sides of the loop and is 0.50 m away from the closer side. at an instant when the emf induced in the loop is 2.0 v, what is the time rate of change of the current in the wire?

Answer 1

The time rate of change of current in the loop is 2 A/s.

The time rate of change of current in the loop.

Faraday's Law states that the electromotive force (emf) induced in a loop is equal to the negative rate of change of magnetic flux through the loop. Mathematically, it is expressed as:

emf = -N * (dΦ/dt)

The loop has a single turn (N = 1), and you are given the induced emf (emf = 2 V).

By rearranging the formula to solve for (dI/dt) (the time rate of change of current),

(dI/dt) = emf / N

(dI/dt) = 2 A/s.

Shown in the following figure is a long, straight wire and a single-turn rectangular loop, both of which lie in the plane of the page. The wire is parallel to the long sides of the loop and is
`$0.50 \mathrm{~m}$` away from the closer side. At an instant when the emf induced in the loop is
`$1.7 \mathrm{~V}$` what is the time rate of change of the current in the wire

Therefore, the time rate of change of current in the loop is 2 A/s.

Answer 2

The time rate of change of current in the loop is determined as 2 A/s.

How to calculate the time rate of change of current?

The time rate of change of current in the loop is calculated by applying the following formula as follows;

emf = NdI/dt

where;

• N is the number of turns of the loop
• dI/dt is the time rate of change of current in the loop

The given parameters include;

number of turns, N = 1

induced emf, emf = 2 V

The time rate of change of current in the loop is calculated as;

dI/dt = emf / N

dI/dt = ( 2 V ) / 1

dI/dt = 2 A/s