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ozzi
3 years ago
14

A long solenoid has a radius and length of 2.30 cm and 25 cm respectively, has 580 turns. If the current in the solenoid is incr

easing at the rate of 17.8 A/s, what is the magnitude of the induced emf produced.
Physics
1 answer:
nadya68 [22]3 years ago
3 0

Answer:

Induced emf is equal to 0.05 volt

Explanation:

We have given length of the solenoid l = 25 cm = 0.25 m

Radius of the solenoid r = 2.3 cm = 0.023 m

Area of the solenoid A=\pi r^2=3.14\times 0.023^2=0.00166m^2

Number of turns N = 580

Inductance of the solenoid L=\frac{\mu _0N^2A}{l}=\frac{4\pi \times 10^{-7\times 580^2\times 0.00166}}{0.25}=0.0028H

Rate of change of current \frac{di}{dt}=17.8A/sec

Induced emf in inductor  is equal to e=L\frac{di}{dt}=0.0028\times 17.8=0.05volt

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gregori [183]

Answer:

a) v=6.898\times 10^{6}\ m.s^{-1} is the speed of each proton

b) F_c=3.97\times 10^{-13}\ N

Explanation:

Given:

radius of path of motion, r=0.2\ m

we know charge on protons, q=1.6\times 10^{-19}\ C

magnetic field strength, B=0.36\ T

we've mass of proton, m=1.67\times 10^{-27}\ kg

a)

From the equivalence of magnetic force and the centripetal force on the proton:

F_B=F_C

q.v.B=\frac{m.v^2}{r}

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where:

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v=6.898\times 10^{6}\ m.s^{-1} is the speed of each proton

b)

Now the centripetal force on each proton:

F_c=m.\frac{v^2}{r}

F_c=1.67\times 10^{-27}\times \frac{(6.898\times 10^6)^2}{0.2}

F_c=3.97\times 10^{-13}\ N

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