Respuesta :
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:
[tex]V = \frac{-BA}{t}[/tex]
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 = [tex]\pi r^2[/tex] = [tex]\pi * 0.2432^2 = 0.186 m^2[/tex]
The magnitude of the Voltage induced will be:
[tex]|V| = |\frac{-0.88 * 0.186}{2.4}| \\\\\\|V| = 0.068 V[/tex]
From Ohm's law, we have that:
[tex]|V| = |I|R[/tex]
=> [tex]|I| = \frac{|V|}{R}[/tex]
[tex]|I| = \frac{0.068}{15}[/tex]
[tex]|I| = 0.0045 A = 4.5 mA[/tex]
The magnitude of the Induced curt 2.4 s after the loop begins expanding is 4.5 mA
Answer:
The magnitude of the induced current is 1.6 mA
Explanation:
The induced emf in the loop is:
[tex]E=-\frac{d\phi }{dt}\\ E=-2\pi r(t)B*0.018[/tex]
[tex]r(t)=r_{o} +0.018t=0.2+0.018t[/tex]
t = 2.4 s
[tex]r(t)=0.2+(0.018*2.4)=0.243m[/tex]
[tex]E=-2\pi *0.243*0.88*0.018=-0.024V=-24mV[/tex]
The induced current is:
[tex]I=\frac{E}{R} =\frac{24}{15} =1.6mA[/tex]
