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 magnetic 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 caused to flow 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 induced emf in a loop is given by Faraday's law of electromagnetic induction:

$$
\epsilon = -\frac{d\phi}{dt}
$$

Where $\epsilon$ is the induced emf, and $\phi$ is the magnetic flux through the loop. Since the magnetic field is uniform and makes an angle of $30^{\circ}$ with the z-axis, the magnetic flux through the loop can be calculated as:

$$
\phi = B \cdot A \cdot \cos{\theta} = B \cdot \pi r^2 \cdot \cos{\theta}
$$

Where $A = \pi r^2$ is the area of the loop, and $\theta$ is the angle between the magnetic field and the surface normal of the loop. Therefore, the induced emf can be written as:

$$
\epsilon = -\frac{dB\pi r^2 \cos{\theta}}{dt}
$$

By Ohm's law, the induced emf causes a current to flow in the loop, given by:

$$
I = \frac{\epsilon}{R}
$$

Where R is the resistance of the loop. In this case, R = 10.0 Ω due to the resistor in the loop. Combining these equations, we have:

$$
I = -\frac{dB\pi r^2 \cos{\theta}}{Rdt}
$$

Substituting the given values, we get:

$$
I = -\frac{(10.0 \mathrm{T} - 1.0\mathrm{T})\pi(0.50\mathrm{m})^2 \cos{30^{\circ}}}{(10.0\mathrm{\Omega})(4.0\mathrm{s})} = 0.15303932 \mathrm{A}
$$

Therefore, the magnitude of the current that will be caused to flow in the loop is 0.153 A, which is closest to option 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 $0.50 \mathrm {~m}$