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 current 2.4 s after the loop begins expanding. A)0 A B)1.3 * 10^-3 A C)1.6 * 10^-3 A D)8.0 * 10^-4 A E)2.1 * 10^-3 A

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The magnitude of the induced current is given by:$$I = \frac{\varepsilon}{R}$$where:* I is the induced current* ε is the induced emf* R is the resistance of the loopThe induced emf is given by:$$\varepsilon = -\frac{d\Phi_B}{dt}$$where:* ε is the induced emf* Φ_B is the magnetic flux through the loopThe magnetic flux through the loop is given by:$$\Phi_B = BA$$where:* B is the magnetic field strength* A is the area of the loopThe area of the loop is given by:$$A = \pi r^2$$where:* A is the area of the loop* r is the radius of the loopSubstituting all of these equations into the equation for the induced current, we get:$$I = -\frac{B\pi r^2}{R}\frac{dr}{dt}$$Plugging in the given values, we get:$$I = -\frac{(0.880 ext{ T})\pi (0.200 ext{ m})^2}{15 \Omega}\frac{0.018 ext{ m}}{1 ext{ s}}$$$$I = -1.6 imes 10^{-3} ext{ A}$$The negative sign indicates that the induced current is flowing in the opposite direction of the magnetic field. Therefore, the magnitude of the induced current is 1.6 * 10^-3 A.

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

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