Question

An elastic circular conducting loop expands at a constant rate in time. The radius is initially 0.200 m. but then begins to increase 0.018 m every second. The loop has a constant resistance of 15 Ohm and Is placed in a region of uniform (same everywhere) magnetic field of 0.880 T, perpendicular to the plane of the loop. Calculate the magnitude of the Induced curt 2.4 s after the loop begins expanding.

Answer 1

Answer:

The magnitude of the induced current is 1.6 mA

Explanation:

The induced emf in the loop is:

$E=-\frac{d\phi }{dt}\\ E=-2\pi r(t)B*0.018$

$r(t)=r_{o} +0.018t=0.2+0.018t$

t = 2.4 s

$r(t)=0.2+(0.018*2.4)=0.243m$

$E=-2\pi *0.243*0.88*0.018=-0.024V=-24mV$

The induced current is:

$I=\frac{E}{R} =\frac{24}{15} =1.6mA$

Answer 2

Answer:

4.5 mA

Explanation:

Parameters given:

Initial radius of coil = 0.2 m

Rate of expansion of coil = 0.018 m/s

Resistance, R = 15 ohms

Magnetic field, B = 0.88 T

Time taken to expand, t = 2.4 s

To find the induced current, we have to first find the induced EMF. This can be gotten by using the formula below:

$V = \frac{-BA}{t}$

where A = area of coil

To find the area, we need to find the radius of the coil 2.4 seconds after it started expanding.

After 2.4s, the coil had expanded to become:

r = 0.2 + (0.018 * 2.4) = 0.2 + 0.0432 = 0.2432 m

Area = $\pi r^2$ = $\pi * 0.2432^2 = 0.186 m^2$

The magnitude of the Voltage induced will be:

$|V| = |\frac{-0.88 * 0.186}{2.4}| \\\\\\|V| = 0.068 V$

From Ohm's law, we have that:

$|V| = |I|R$

=> $|I| = \frac{|V|}{R}$

$|I| = \frac{0.068}{15}$

$|I| = 0.0045 A = 4.5 mA$

The magnitude of the Induced curt 2.4 s after the loop begins expanding is 4.5 mA