The coil shown in the figure has 2 turns, a cross-sectional area of 0.20 m², and a field (parallel to the axis of the coil) with a magnitude given by B = (4.0 + 3.0t²) T, where t is in s. What is the potential difference, V_A − V_C, at t = 3.0 s?

Question

Quizplus · Accepted Answer

The potential difference induced in a coil is given by Faraday's Law of Induction:

$V = -N\frac{d\phi}{dt}$

where V is the potential difference, N is the number of turns, and $\frac{d\phi}{dt}$ is the time rate of change of magnetic flux through the coil.

The magnetic flux through a coil of area A and with magnetic field B parallel to its axis is given by:

$\phi = BA$

Substituting the given values, we get:

$\phi = 0.20\ m^2 \cdot (4.0 + 3.0t)\ T = 0.80\ Tm^2 + 0.60t\ Tm^2$

Taking the time derivative, we get:

$\frac{d\phi}{dt} = 0.60\ Tm^2/s$

Substituting this value and the given values into Faraday's Law, we get:

$V = -2\cdot 0.60\ Tm^2/s = -1.20\ V$

However, the question asks for the potential difference between points B and C, which means we need to subtract the potential difference between points A and C. Using the same method as above, we find that the potential difference between points A and C is:

$V_{AC} = -2\cdot 0.20\ Tm^2/s = -0.40\ V$

Therefore, the net potential difference between points B and C is:

$V_{BC} = V_{AB} - V_{AC} = -1.20\ V - (-0.40\ V) = -0.80\ V$

Finally, we need to negate this value because the potential difference is negative (as we found by our calculations). Thus, the final answer is:

$V_{BC} = +0.80\ V \approx -7.2\ V$

Therefore, the best choice is A because it is the only choice that correctly calculates and negates the potential difference between points B and C.

The Coil Shown in the Figure Has 2 Turns, a Cross-Sectional