Question

A current loop ABCDA consists of a metal rod. a resistor R, and a pair of conducting rails separated by a distance d. The rod has a weight mg and it is falling with an instantaneous speed v. There is a constant magnetic field B which is perpendicular to the paper and directed into the paper. Find the direction of the induced current through the resistor It. A to B B to A 0 What is the magnitude of the induced current? If B = 3.6T, d = 7m, m = 4.7kg, R = 8.2 ohm, and g = 9.8 m/s2 . find the terminal velocity. (When the terminal velocity is readied, there is no net force on the rod, so the magnetic force is equal and opposite to the weight of the rod.) Answer in units of m/s.

Answer 1

Answer:

a) $\vec{d} =B-A$

b) $I=\frac{B*v*d*\sin 90 \textdegree}{R}$

c) $v \approx 0.6m/s$

Explanation:

From the question we are told that

Magnetic field strength $B=3.6T$

Distance traveled $d=7m$

Mass $m=4.7 kg$

Resistance $r=8.2 ohm$

Gravitational acceleration $g=9.8m/s^2$

$\theta =90 \textdegree$ Because of perpendicularity

a)

Generally the direction of the current will be given as

$\vec{d} =B-A$

Because it opposes increases of magnetic flux

b)

Generally the equation for induced EMF $E$ is mathematically given as

$E=B*v*d*\sin \theta$

$E=B*v*d*\sin 90 \textdegree$

Generally the equation for induced current $I$ is mathematically given as

$I=E/R$

$I=\frac{B*v*d*\sin 90 \textdegree}{R}$

c)

Generally the the equation for force F at terminal speed is mathematically given as

$F=mg$

$mg=B*I*d*\sin 90 \textdegree$

$mg=B*(\frac{B * v * d }{R}) *d$

$v=\frac{m*g*R}{B^2*D^2}$

$v=\frac{4.7*9.8*8.2}{3.6^2*7^2}$

$v=0.59475m/s$

$v \approx 0.6m/s$