What is the magnitude of the induced current in the coil

[kbyws37]

The component of the external magnetic field along the central axis of a 50−turn coil of radius 5.0 cm increases from 0 to 1.0 T in 4.3 s. (a) If the resistance of the coil is 2.9 Ω, what is the magnitude of the induced current in the coil? (b) What is the direction of the current if the axial component of the field points away from the viewer?


I am trouble starting this problem.
Which equation am I suppose to start with.
I am trying to use
emf = vBL
but I don't know what v would be.
i tried to use Faraday's law equation but i didn't have all of the exponents.

 

[hage567]

You are not applying Faraday's law correctly for this particular question. For (a), you want to consider the rate of change of magnetic flux through the coil. Think of the coil as 50 circular loops. Think of the area of a loop.

 

[m2003]

As a scientist, it is important to have a strong understanding of the fundamental principles and equations related to electromagnetism. In this scenario, we are dealing with Faraday's law, which states that the induced emf (electromotive force) in a closed loop is equal to the negative rate of change of magnetic flux through the loop. This can be written as:

emf = -N(dΦ/dt)

Where N is the number of turns in the coil, Φ is the magnetic flux, and dt is the change in time. In this case, we are given the number of turns (N = 50) and the change in magnetic field (ΔB = 1.0 T). We also know the time it takes for this change to occur (Δt = 4.3 s). Therefore, we can use these values to calculate the change in magnetic flux through the coil.

Φ = BA = ΔBπr^2

Where B is the magnetic field, A is the cross-sectional area of the coil, and r is the radius. Plugging in the given values, we get:

Φ = (1.0 T)(π)(0.05 m)^2 = 7.85 x 10^-4 Tm^2

Now, we can go back to Faraday's law and plug in the values we have calculated:

emf = -(50)(7.85 x 10^-4 Tm^2)/(4.3 s) = -9.14 x 10^-4 V

Since we know the resistance of the coil (R = 2.9 Ω), we can use Ohm's law to calculate the induced current (I = V/R):

I = (-9.14 x 10^-4 V)/(2.9 Ω) = -3.15 x 10^-4 A

Therefore, the magnitude of the induced current in the coil is 3.15 x 10^-4 A.

As for the direction of the current, we can use the right-hand rule to determine that it will be counterclockwise when viewed from the end of the coil where the axial component of the field points away from the viewer. This is because the change in magnetic field is causing a change in flux, which induces a current in the opposite direction according to Lenz's law.

In conclusion, as a scientist, it is important to have a