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Zinaida [17]
3 years ago
11

Consider the uniform electric field \vec{E} =(4.00~\hat{j}+3.00~\hat{k})\times 10^3~\text{N/C} ​E ​⃗ ​​ =(4.00 ​j ​^ ​​ +3.00 ​k

​^ ​​ )×10 ​3 ​​ N/C. What is its electric flux through a circular area of radius 2.66 m that lies in the xy-plane?
Physics
1 answer:
Tema [17]3 years ago
6 0

Answer:

Electric flux, \phi=6.668\times 10^4\ Nm^2/C

Explanation:

It is given that,

Electric field, E=(4j+3k)\times 10^3\ N/C

We need to find the electric flux through a circular area of radius 2.66 m that lies in the xy-plane. A=Ak

The electric flux is given by :

\phi=E{\cdot}A

\phi=(4j+3k)\times 10^3{\cdot}Ak

Since, k.k=i.i=j.j = 1

So,

\phi=3\times 10^3\times \pi\times (2.66)^2\ k

\phi=6.668\times 10^4\ Nm^2/C

So, the electric flux through a circular area is \phi=6.668\times 10^4\ Nm^2/C.Hence, this is the required solution.

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What is the charge of an atom with 21 protons and 6 electrons
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Scandium with an ion charge of +3

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Read 2 more answers
When a magnetic field is first turned on, the flux through a 20-turn loop varies with time according to Φm=5.0t2−2.0t, where Φm
goldfiish [28.3K]

Answer:

A) ( - 200t + 40 ) volts

B)  b) anticlockwise ,  c) anticlockwise , d) clockwise ,  e) clockwise

Explanation:

Given data:

magnetic flux (Φm) = 5.0t^2 − 2.0t

number of turns = 20

<u>a) determine induced emf </u>

E = - N \frac{d\beta }{dt}

  =  - N ( 10t - 2 ) = - 20 ( 10t - 2 )

  =  - 200t + 40  volts

<u>b) Determine direction of induced current </u>

i) at t = 0

 E = - 0 + 40  ( anticlockwise direction )

ii) at t = 0.10

E = -20 + 40 =  20 ( anticlockwise direction )

iii) at t = 1

E = - 200 + 40 = - 160 ( clockwise direction)

iv) at t = 2

E = -400 + 40 =  - 360 ( clockwise direction )

8 0
2 years ago
A 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m,
Tems11 [23]

Answer:

3.6 KJ

Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

The workdone = the energy.

There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )

P.E = mgh

P.E = 70 × 9.8 × 1.6

P.E = 1097.6 J

P.E = 1.098 KJ

K.E = 1/2mv^2

K.E = 1/2 × 70 × 8.5^2

K.E = 2528.75 J

K.E = 2.529 KJ

The non conservative workdone = K.E + P.E

Work done = 1.098 + 2.529

Work done = 3.63 KJ

Therefore, the non conservative workdone is 3.6 KJ approximately

5 0
2 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the lo
Zepler [3.9K]

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

\phi = BAcos \theta

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb

Hence the magnetic flux Φ through the loop is 0.5849Weber

5 0
2 years ago
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