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Dafna1 [17]
1 year ago
15

two circular loops of wire, each containing a single turn, have the same radius of 4.0 cm and a common center. the planes of the

loops are perpendicular. each carries a current of 1.7 a. what is the magnitude of the net magnetic field at the common center?
Physics
1 answer:
REY [17]1 year ago
3 0

The magnetic field at center of circular loops of wire is 3.78 x 10¯⁵ T.

We need to know about the magnetic field at the center of circular loops of wire to solve this problem. The magnetic field at the center can be determined as

B = μ₀ . I / 2r

where B is magnetic field, μ₀ is vacuum permeability (4π×10¯⁷ H/m), I is the current and r is radius.

From the question above, we know that:

r = 4 cm = 0.04 m

I = 1.7 A

By substituting the parameter, we get

B = μ₀ . I / 2r

B = 4π×10¯⁷ . 1.7 / (2.0.04)

B = 2.67 x 10¯⁵ T

Due to the perpendicular plane of loops, the total magnetic field at center will be

Btotal = √(2(B²))

Btotal = √(2(2.67 x 10¯⁵²))

Btotal = 3.78 x 10¯⁵ T

Find more on magnetic field at: brainly.com/question/7802337

#SPJ4

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A vaulter holds a 26.90-N pole in equilibrium by exerting an upward force U with her leading hand and a downward force D with he
ivolga24 [154]

Answer:

Check attachment for better understanding

Explanation:

Given that

a= 0.75m

b=1.31m

c= 2.2m

Weight of pole is 26.90

Then, Fg = Weight = 26.90

Using Equilibrium of forces

ΣFy = 0

U — D — Fg = 0

U — D = Fg

U — D = 26.9

To calculate U,

We will take moment about point A.

ΣMa = 0

Let the clockwise moment be positive and anti-clockwise be negative

Fg(a+b) — U(a) = 0

26.9(1.31+0.75) —0.75U = 0

26.9(2.06) = 0.75U

0.75U = 55.414

U = 55.414/0.75

U = 73.89 N

To calculate D,

U — D = 26.9

73.89—D =26.9

73.89—26.9 = D

D = 46.99N

3 0
3 years ago
Read 2 more answers
A 20×10⁹charge is moved between two points A andB that are 30mm apart and have an electric potential difference of 600v between
Ulleksa [173]

Answer:

90x20=1800

Explanation:

just multiply 10 & 9 and then mutiply 90x20 or 20x90

4 0
2 years ago
I dont know why people find 21 and 17 38 funny plz explain
ElenaW [278]
Not sure about 21, but the 1738 is a turned used by a rapper named Fetty Wap in his song called Trap Queen to represent the "Remy Boyz 1738" which they named themselves that after the finest 1738 Remy Martin liquor. 
Hope that helps you! :)
3 0
3 years ago
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An investigator places a sample 1.0 cm from a wire carrying a large current; the strength of the magnetic field has a particular
AURORKA [14]

Answer:

she must increase the current by factor of 7

Explanation:

The magnetic field produced by a steady current flowing in a very long straight wire encircles the wire.In order to solve the question, we use this formula,

B= μo I/(2πr)

where,

'μo'  represents permeability of free space i.e 4π*10-7 N/A2

B=magnetic field

I= current

r=radius

->When r= 1cm=> 0.01m

B1 = μo I_{1/(2π x 0.01)

->when r=7cm =>0.07m

B2 = μo I_{2}/(2π x 0.07)

Now equating both of the magnetic fields, we have

B1= B2

μo I_{1/(2π x 0.01)= μo I_{2}/(2π x 0.07)

I_{1/  I_{2}= 0.01/0.07

I_{1/  I_{2}= 1/ 7

Therefore, she must increase the current by factor of 7

5 0
3 years ago
a stone is thrown horizonttaly from a cliff of a hill with an initial velocity of 30m/s it hits the ground at a horizontal dista
ELEN [110]

Answer:

a) Time = 2.67 s

b) Height = 35.0 m

Explanation:

a) The time of flight can be found using the following equation:

x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}   (1)

Where:

x_{f}: is the final position in the horizontal direction = 80 m

x_{0}: is the initial position in the horizontal direction = 0

v_{0_{x}}: is the initial velocity in the horizontal direction = 30 m/s

a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)

t: is the time =?  

By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:  

t = \frac{x_{f}}{v_{0}} = \frac{80 m}{30 m/s} = 2.67 s

b) The height of the hill is given by:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final position in the vertical direction = 0

y_{0}: is the initial position in the vertical direction =?

v_{0_{y}}: is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)            

g: is the acceleration due to gravity = 9.81 m/s²

Hence, the height of the hill is:

y_{0} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(2.67 s)^{2} = 35.0 m  

I hope it helps you!

5 0
3 years ago
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