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Bogdan [553]
3 years ago
9

A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

i of amplitude 3.08 A and angular frequency 226 rad/s. The central axis of the loop and solenoid coincide. What is the amplitude of the emf induced in the loop?
Physics
1 answer:
Tema [17]3 years ago
8 0

Answer:410.90\times 10^{-6} V

Explanation:

Given

A=5.5 mm^2

n=919 turns/cm\approx 85400 turns/m

\omega =226 rad/s

According to the Faraday law of induction, induced emf is given by

E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}

Magnetic field \phi _B=B\cdot A

B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )

B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)

B=0.3305\sin (226t)

\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}

\phi _{B}=1.818\times 10^{-6}\sin (226t)

E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}

E=-1.818\times 10^{-6}\times 226\cos (226t)

E=-410.90\times 10^{-6}\cos (226t)

Amplitude of EMF induced=410.90\times 10^{-6} V

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The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is
astraxan [27]

Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer:

The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

Finally, we determine the magnitude of this field;

E = \frac{T}{P*sin(\theta)}=  \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C

E = 826 N/C (in three significant figures)

Therefore, the magnitude of this field is 826 N/C

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In Thomson’s experiment, why was the glowing beam repelled by a negatively charged plate?
Svetllana [295]

The glowing beam was repelled by a negatively charged plate because they were negatively charged

<h3>What are the nature of charges?</h3>

The nature of charges refers to the properties of charges.

There are two types of charges:

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The law of electricity states that opposite charges attract whereas like charges repel.

Therefor, in Thomson’s experiment, the glowing beam was repelled by a negatively charged plate because they were negatively charged

In conclusion, like charges repel while opposite charges attract.

Learn more about charges at: brainly.com/question/12781208

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The car can travel 22.5km in 0.5 hours
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