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3 years ago

NEED HELP

Physics

1 answer:

aniked [119]3 years ago

5 0

Answer:

I = 0.25 [amp]

Explanation:

To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance.

V = I*R

where:

V = voltage [Volt]

I = amperage or current [amp]

R = resistance [ohm]

Since all resistors are connected in series, the total resistance will be equal to the arithmetic sum of all resistors.

Rt = 2 + 8 + 14

Rt = 24 [ohm]

Now clearing I for amperage

I = V/Rt

I = 6/24

I = 0.25 [amp].

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A solenoid coil with 22 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 25.0 cm long an

LUCKY_DIMON [66]

Answer:

a) 1.34*10^-8 W

b) 1.18*10^-5 H

c) 20mV

Explanation:

a) To find the average magnetic flux trough the inner solenoid you the following formula:

$\Phi_B=BA=\mu_oNIA$

mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A

N: turns of the solenoid = 340

I: current of the inner solenoid = 0.100A

A: area of the inner solenoid = pi*r^2

r: radius of the inner solenoid = 2.00cm/2=1.00cm=10^-2m

You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:

$A=\pi(10^{-2}m)^2=3.141510^{-4}m^2\\\Phi_B=(4\pi*10^{-7}T/A)(340)(0.100A)(3.1415*10^{-4}m^2)=1.34*10^{-8}W\\$

the magnetic flux is 1.34*10^{-8}W

b) the mutual inductance is given by:

$M=\mu_o N_1 N_2 \frac{A_2}{l}$

N1: turns of the outer solenoid = 22

N2: turns of the inner solenoid

A_2: area of the inner solenoid

l: length of the solenoids = 25.0cm=0.25m

by replacing all these values you obtain:

$M=(4\pi*10^{-7}T/A)(340)(22)\frac{3.14*10^{-4}m^2}{0.25m}=1.18*10^{-5}H$

the mutual inductance is 1.18*10^{-5}H

c) the emf induced can be computed by using the mutual inductance and the change in the current of the inner solenoid:

$\epsilon_1=M\frac{dI_2}{dt}$

by replacing you obtain:

$\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV$

the emf is 20mV

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