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Vilka [71]
3 years ago
9

The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the w

ire represent the direction of current). For the following wires, which all also carry current , indicate the magnitude (in mT) and direction of the magnetic field at the center (red point) of each configuration.

Physics
1 answer:
Citrus2011 [14]3 years ago
3 0

Complete Question

 The complete question is shown on the first uploaded image

Answer:

The magnetic field is B_{net} = \frac{1}{4}  * mT

And the direction is  -\r k

Explanation:

      From the question we are told that

                 The magnetic field at the center is B = 1mT

Generally magnetic field is mathematically represented as

              B = \frac{\mu_o I}{2R}

We are told that it is equal to 1mT

So

                B = \frac{\mu_o I}{2R} = 1mT

From the first diagram we see that the effect of the current flowing in the circular loop is  (i.e the magnetic field generated)

                         \frac{\mu_o I}{2R} = 1mT

 This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

                   B_1 = \frac{1}{2} \frac{\mu_o I}{2R}

and  for the larger semi-circular loop  is

                 B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction

    So the net magnetic field would be

                   B_{net} = B_1 - B_2

                        =  \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}

                        =\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}

                        =\frac{\mu_o I}{8R}

                        = \frac{1}{4} \frac{\mu_o I}{2R}

Recall  \frac{\mu_o I}{2R} = 1mT

    So  

             B_{net} = \frac{1}{4}  * mT

Using the Right-hand rule we see that the direction is into the page which is -k

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3 years ago
A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'
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Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom v=4.22 m/s

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As Energy is conserved therefore Energy at top is equal to energy at bottom

E_T=E_B

\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}  ,where v_0 is the velocity at bottom

\frac{v^2}{2}+gh=\frac{v_0^2}{2}

v_0^2=v^2+2gh

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v=\sqrt{v_0^2-2gh}

v=\sqrt{4.22^2-2\times 9.8\times 0.286}

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7 0
3 years ago
A student is bicycling north along Main Street to school. Another student is timing the bicycling student in order to determine
MatroZZZ [7]

The average velocity is -4.17 m/s

Explanation:

The average velocity of a body is given by:

v=\frac{d}{t}

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d is the displacement of the body

t is the time elapsed

For the student in this problem, we have:

Initial position: x_i = 450 m

Final position: x_f = 200 m

So the displacement is

d=x_f -x_i = 200 - 450 = -250 m

The time elapsed is

t = 60 s

Therefore, the average velocity is

v=\frac{-250}{60}=-4.17 m/s

Where the negative sign means the student is moving towards the origin.

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brainly.com/question/8893949

brainly.com/question/5063905

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8 0
3 years ago
You run a race with your friend. At first you each have the same kinetic energy, but then you find that she is beating you. When
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Answer:

52.49 Kg

Explanation:

Let m1 and v1 denote your mass and velocity respectively

Let m2 and v2 denote your friends mass and velocity respectively

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KE= 0.5mv^{2}  

Since your kinetic energies are the same hence

0.5m1(v1)^{2}=0.5m2(v2)^{2}

m1(v1)^{2}=m2(v2)^{2} and making m2 the subject then  

m2=\frac { m1(v1)^{2}}{(v2)^{2}}

Since v2 is v1+0.28v1=1.28v1

Substituting m1 for 86 Kg

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3 0
3 years ago
g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located
Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is:  P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P  /RT ⋯ ⋯( I )

Here  

m  is the mass of the air,  

v  is the volume of the air,  

P  is the atmospheric pressure,  

T  is the standard temperature at the atmospheric pressure and  

R  is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101  kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here  A  is the area of the windmill

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A = π /4  d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94  k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94  k g / s ( (6m/s)²/2 )

P w = 160144.92 W  × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0
3 years ago
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