Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
The answer is B. Bye because B those study speed.
To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,
Here,
= Angle between areal vector and magnetic field direction.
According to Faraday's law, induced emf in the loop is,
At time 
, Induced emf is,
Therefore the magnitude of the induced emf is 10.9V
Answer:
1250 J
Explanation:
Work is said to be done when a force causes an object to move over a distance. The amount of work done (W) is calculated by multiplying the force by the distance traveled.
That is;
W = F × d
Where;
W = work done (J or N/m)
F = force (N)
d = distance (m)
Based on the information provided in this question, F = 5000N, d = 0.25m
Hence;
W = F × d
W = 5000 × 0.25
W = 1250J
Therefore, 1250Joules of work is done by the jack.
