a conducting loop consisting of a half circle of radius r=0.2m and three straight sections the half circle lies in a uniform magnetic field B that is directed out of the page the field magnitude is given by B=(4t^2 + 2t + 3)T .w/r t is in second. an ideal battery with emf=2v is connected to the loop the resistance of the loop is 2 ohm.
a)what is the magnitude of the induced emf at t=10sec
b)what is the current in the loop at t=10sec
To find the magnitude of the induced EMF and the current in the loop at t = 10 sec, we need to use Faraday's law of electromagnetic induction and Ohm's law. Here's how you can solve both parts of the question:
a) Magnitude of the induced EMF at t = 10 sec:
According to Faraday's law, the induced EMF (ε) in a loop is equal to the rate of change of magnetic flux through the loop. The magnetic flux (Φ) through the loop is given by the product of the magnetic field (B) and the enclosed area (A).
In this case, the loop consists of a half-circle and three straight sections. The half-circle is the enclosed area. The radius of the half-circle is given as r = 0.2 m. The magnetic field B is given by B = (4t^2 + 2t + 3) T.
To find the magnetic flux, we need to integrate the magnetic field vector B over the enclosed area. The formula for calculating the magnetic flux through a curved surface is:
Φ = ∫ B⋅dA
The integral of B with respect to dA over a half-circle can be simplified by using the relationship between the radius (r) and the angle (θ):
r = R⋅θ
dA = r⋅dl = R⋅θ⋅R⋅dθ = R^2⋅θ⋅dθ
Therefore, the enclosed area can be written as:
A = ∫ R^2⋅θ⋅dθ
Substituting this value for A, we can now calculate the induced EMF at t = 10 sec:
ε = -dΦ/dt = -d/dt ∫ R^2⋅θ⋅dθ
You'll need to evaluate this derivative using the given magnetic field expression, B = (4t^2 + 2t + 3) T.
b) Current in the loop at t = 10 sec:
To find the current in the loop at t = 10 sec, we can use Ohm's law, which states that the current (I) in a conductor is the ratio of the voltage (V) across the conductor to its resistance (R):
I = V / R
In this case, the voltage (V) across the conductor is the induced EMF (ε) calculated in part a, and the resistance (R) of the loop is given as 2 Ω.
Therefore, to find the current at t = 10 sec, substitute the value of ε from part a and the resistance value into the Ohm's law equation.