An elastic circular conducting loop expands at a constant rate in time.The radius is initially 0.200 m,but then begins to increase 0.018 m every second.The loop has a constant resistance of 15 $\varOmega$and is placed in a region of uniform (same everywhere)magnetic field of 0.880 T,perpendicular to the plane of the loop.Calculate the magnitude of the induced EMF 2.4 s after the loop begins expanding.
A)31.2 mV
B)24.2 mV
C)22.3 mV
D)28.0 mV
E)19.9 mV

Question

Quizplus · Accepted Answer

The induced EMF ($ \varepsilon $) in a loop due to its expansion in a magnetic field can be calculated using Faraday's law of electromagnetic induction, which states that the induced EMF is equal to the rate of change of the magnetic flux through the loop. The magnetic flux ($ \Phi $) through the loop is given by $ \Phi = B \cdot A $, where $ B $ is the magnetic field strength and $ A $ is the area of the loop. For a circular loop, $ A = \pi r^2 $, where $ r $ is the radius of the loop.Given that the radius increases at a constant rate, $ \frac{dr}{dt} = 0.018 $ m/s, and the initial radius is $ r_0 = 0.200 $ m, the radius at any time $ t $ seconds after the expansion begins is $ r = r_0 + 0.018t $.At $ t = 2.4 $ s, the radius is $ r = 0.200 + 0.018 \cdot 2.4 = 0.2432 $ m.The area of the loop at this time is $ A = \pi r^2 = \pi (0.2432)^2 $ m$^2$.The rate of change of the area, $ \frac{dA}{dt} $, is important for calculating the induced EMF. Since $ A = \pi r^2 $, $ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} $.Substituting $ r = 0.2432 $ m and $ \frac{dr}{dt} = 0.018 $ m/s, we get $ \frac{dA}{dt} = 2\pi (0.2432)(0.018) $.The induced EMF is then $ \varepsilon = B \frac{dA}{dt} = 0.880 \cdot 2\pi (0.2432)(0.018) $.Calculating this gives $ \varepsilon \approx 24.2 $ mV.

An Elastic Circular Conducting Loop Expands at a Constant Rate Ω\varOmegaΩ

An Elastic Circular Conducting Loop Expands at a Constant Rate $\varOmega$