A single-turn loop of wire, having a resistance of $8.00 \Omega$ and a cross-sectional area $200 \mathrm {~cm} ^ { 2 }$, is perpendicular to a uniform magnetic field that increases steadily from $0.200 \mathrm {~T}$ to $2.800 \mathrm {~T}$ in $2.20$ seconds. What is the magnitude of the induced current in the loop?
A) $0 \mathrm {~A}$
B) $2.95 \mathrm {~mA}$
C) $3.18 \mathrm {~mA}$
D) $2.95 \mathrm {~A}$
E) $3.18 \mathrm {~A}$

Question

Quizplus · Accepted Answer

The induced current can be calculated using Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) in a loop is equal to the rate of change of magnetic flux through the loop. The formula for EMF ($E$) is $E = -\Delta \Phi / \Delta t$, where $\Delta \Phi$ is the change in magnetic flux and $\Delta t$ is the change in time. The magnetic flux ($\Phi$) is given by $\Phi = B \cdot A$, where $B$ is the magnetic field and $A$ is the area through which the field lines pass, perpendicular to the field. The area should be converted to square meters: $200 \mathrm{cm}^2 = 0.02 \mathrm{m}^2$. The change in magnetic flux is $\Delta \Phi = B_f \cdot A - B_i \cdot A = (2.800 - 0.200) \cdot 0.02 = 0.052 \mathrm{Wb}$. The change in time is $\Delta t = 2.20 \mathrm{s}$. Thus, $E = 0.052 \mathrm{Wb} / 2.20 \mathrm{s} = 0.02364 \mathrm{V}$. The induced current ($I$) can then be found using Ohm's law, $I = E / R$, where $R = 8.00 \Omega$. Therefore, $I = 0.02364 \mathrm{V} / 8.00 \Omega = 0.002955 \mathrm{A} = 2.955 \mathrm{mA}$, which rounds to $2.95 \mathrm{mA}$.

A Single-Turn Loop of Wire, Having a Resistance Of 8.00Ω8.00 \Omega8.00Ω

A Single-Turn Loop of Wire, Having a Resistance Of $8.00 \Omega$