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stira [4]
2 years ago
10

When the current through a circular loop is 5.7 A, the magnetic field at its center is 3.9 ✕ 10−4 T. What is the radius (in m) o

f the loop?
Physics
1 answer:
borishaifa [10]2 years ago
8 0

Radius of the circular loop is 0.0091m.

<h3>What is magnetic field?</h3>

Magnetic field is the area around a magnet where the magnetism influence is felt .

<h3>What is the magnetic field at the centre of a circular loop?</h3>
  • The formula for magnetic field at the centre of a loop is

B =(μ)I/2r

  • where B= Magnetic field at the centre of a circular loop

μ= Magnetic permeability =4(π)*10^(-7)

I= current flowing through the loop

r= radius of the loop

  • Thus, radius of the loop =(4(π)×10^(-7)×5.7)/(2×3.9×10^(-4))

=0.0091m

Thus, we can conclude that the radius of the loop is 0.0091m .

Learn more about magnetic field here :

brainly.com/question/24761394

#SPJ1

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What is the current in a 160V circuit if the resistance is 2Ω? V= I= R =
sveta [45]

Answer:

80 amperes

Explanation:

Current in the circuit = ?

Voltage in the circuit = 160 Volts

Resistance = 2 Ω

Voltage = Current x Resistance

V = IR

160V = I x 2 Ω

I = 160V / 2 Ω

I = 80 Amperes

Therefore the current in the circuit is 80 amperes :)

8 0
3 years ago
Suzette had prepared the graph below to add to her lab
Charra [1.4K]

Answer:

A title

Explanation:

Because this is middle school.

4 0
3 years ago
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A certain unfiltered full-wave rectifier with 120 V, 60 Hz input produces an output with a peak of 15 V. When a capacitor-input
KATRIN_1 [288]

Answer:

The peak-to-peak ripple voltage = 2V

Explanation:

120V and 60 Hz is the input of an unfiltered full-wave rectifier

Peak value of  output voltage = 15V

load connected = 1.0kV

dc output voltage = 14V

dc value of the output voltage of capacitor-input filter

where

V(dc value of output voltage) represent V₀

V(peak value of output voltage) represent V₁

V₀ = 1 - ( \frac{1}{2fRC})V₁

make C the subject of formula

V₀/V₁ = 1 - (1 / 2fRC)

1 / 2fRC = 1 - (v₀/V₁)

C = 2fR ((1 - (v₀/V₁))⁻¹

Substitute  for,

f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V

C = 2 * 240 * 1 (( 1 - (14/15))⁻¹

C = 62.2μf

The peak-to-peak ripple voltage

= (1 / fRC)V₁

= 1 /  ( (120 * 1 * 62.2) )15V

= 2V

The peak-to-peak ripple voltage = 2V

3 0
4 years ago
If the distance to a point source of sound is doubled, by what multiplicative factor does the intensity change?
alina1380 [7]

If the distance to a point source of sound is doubled, by a multiplicative factor of 4, the intensity changes.

Intensity of sound is the sound which is perpendicular to sound wave propogation per unit area. It is dependent on the Surface of source sound.

Intensity is the Power per unit area. Its SI unit is Watt/m².

As we move away from a source of sound, the sound starts to diminish. This is due to the decreasing sound intensity with distance.

It can also be understood by the fact that on increasing distance, the Power radiated by the source spreads over a larger area. Hence, the Intensity decreases gradually.

Since, Intensity is proportional to the square of the distance.

Hence, on doubling the distance, Intensity reduces to one fourth of the initial intensity or reduces by a multiplicative factor of 4.

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8 0
2 years ago
What are some physical properties of matter
vesna_86 [32]
Maybe this will help you in a way,<span> some physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, and many others.



  - hope this helps you well .</span>
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