A circular conducting loop with a radius of $0.50 \mathrm {~m}$ and a small gap filled with a $10.0 - \Omega$ resistor is oriented in the $x y - p l a n e .$ If a uniform $\vec { B }$ field of $1.0 \mathrm {~T}$ making an angle of $30 ^ { \circ }$ with the $\text { z-axis, }$ increases to $10.0 \mathrm {~T} ,$ in $4.0 \mathrm {~s}$ what is the magnitude of the current that will be induced in the loop if it has negligible resistance?
A)0.01530393 A
B)0.08835729 A
C)0.15303932 A
D)0 A

Question

Quizplus · Accepted Answer

The magnitude of 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 magnetic flux $\Phi_B$ through the loop is given by $\Phi_B = BA\cos(\theta)$, where $B$ is the magnetic field strength, $A$ is the area of the loop, and $\theta$ is the angle between the magnetic field and the normal to the plane of the loop. The change in magnetic flux $\Delta \Phi_B$ when $B$ changes from $1.0 \,T$ to $10.0 \,T$ is $\Delta \Phi_B = \Delta B \cdot A \cdot \cos(\theta)$. The area $A$ of the loop is $\pi r^2$, where $r = 0.50 \,m$. The angle $\theta = 30^\circ$, so $\cos(30^\circ) = \sqrt{3}/2$. The change in $B$ is $10.0 \,T - 1.0 \,T = 9.0 \,T$. Thus, $\Delta \Phi_B = 9.0 \cdot \pi \cdot (0.50)^2 \cdot \sqrt{3}/2$. The induced emf ($\epsilon$) is equal to the rate of change of magnetic flux, so $\epsilon = \Delta \Phi_B / \Delta t$, where $\Delta t = 4.0 \,s$. The induced current $I$ can be found using Ohm's Law, $I = \epsilon / R$, where $R = 10.0 \,\Omega$. Substituting the values and calculating gives $I \approx 0.153 \,A$, which matches choice C.

A Circular Conducting Loop with a Radius Of 0.50 m0.50 \mathrm {~m}0.50 m

A Circular Conducting Loop with a Radius Of $0.50 \mathrm {~m}$