A conducting rectangular loop of mass M, resistance R, and dimensions a × b is allowed to fall from rest through a uniform magnetic field which is perpendicular to the plane of the loop. The loop accelerates until it reaches a terminal speed (before the upper end enters the magnetic field). If a = 2.0 m, B = 6.0 T, R = 40 Ω, and M = 0.60 kg, what is the terminal speed?
A) 1.6 m/s
B) 20 m/s
C) 2.2 m/s
D) 26 m/s
E) 5.3 m/s

Question

A conducting rectangular loop of mass M, resistance R, and dimensions a × b is allowed to fall from rest through a uniform magnetic field which is perpendicular to the plane of the loop. The loop accelerates until it reaches a terminal speed (before the upper end enters the magnetic field). If a = 2.0 m, B = 6.0 T, R = 40 Ω, and M = 0.60 kg, what is the terminal speed? 
A) 1.6 m/s 
B) 20 m/s 
C) 2.2 m/s 
D) 26 m/s 
E) 5.3 m/s

Quizplus · Accepted Answer

The force acting on the loop in the magnetic field is given by $F=BIL$. The current in the loop can be found using the formula $V=IR$, where $V$ is the induced EMF due to the motion of the loop in the magnetic field. The EMF can be found using Faraday's law $\epsilon=-\frac{d\phi_B}{dt}$, where $\phi_B$ is the magnetic flux through the loop.

The magnetic flux through the loop is $\phi_B=BA$, where $A$ is the area of the loop. The rate of change of magnetic flux is given by $\frac{d\phi_B}{dt}=B\frac{dA}{dt}$. Since the loop is falling at a constant speed, the rate of change of area is $\frac{dA}{dt}=ab\frac{d}{dt}\left(\frac{h}{a}ight)=\frac{b}{a}v$, where $h$ is the height of the loop above the magnetic field and $v$ is the velocity of the loop. Therefore, $\frac{d\phi_B}{dt}=B\frac{b}{a}v$.

The induced EMF is then $\epsilon=-\frac{d\phi_B}{dt}=-B\frac{b}{a}v$. The current in the loop is $I=\frac{\epsilon}{R}=-\frac{Bb}{aR}v$. The force acting on the loop is $F=BIL=-\frac{B^2ab}{aR}v$. The direction of the force is opposite to the direction of motion, so the acceleration of the loop is $a=\frac{F}{M}=-\frac{B^2ab}{MaR}v$.

The terminal speed is reached when the force due to gravity is equal to the force due to the magnetic field. The force due to gravity is $Mg$, where $g$ is the acceleration due to gravity. Therefore, we have $-\frac{B^2ab}{MaR}v=Mg$.

Solving for $v$, we get $v=-\frac{MgRa}{B^2b}$. Plugging in the given values, we get $v\approx1.6	ext{ m/s}$. Therefore, the answer is A.

A Conducting Rectangular Loop of Mass M, Resistance R, and Dimensions