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horsena [70]
3 years ago
10

A circular loop has radius R and carries current I2 in a clockwise direction. The center of the loop is a distance D above a lon

g, straight wire.
What is the magnitude of the current I1 in the wire if the magnetic field at the center loop is zero? Express your answer in terms of the variables I2, R, D, and appropriate constants (μ0 and π).
Physics
1 answer:
romanna [79]3 years ago
5 0

Answer:

I_{1} = (πDI_{2})/R

Explanation:

If we define the magnitude of the field as B, then we have:

Total magnitude of the field B_{t} = magnitude of the field B_loop + magnitude of the field B_wire. The total magnitude is equivalent to zero. Therefore, the field B_loop has an inward direction while the field B_wire has an outward direction.

B_loop = (μ0)*(I_{2})/2*R

B_wire = (μ0)*(I_{1})/2*π*D

Thus:

B_loop = B_wire at the center of the loop.

(μ0)*(I_{2})/2*R = (μ0)*(I_{1})/2*π*D

I_{1} = (πDI_{2})/R

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2. A 1.54 kΩ resistor is connected to an AC voltage source with an rms voltage of 240 V.
svp [43]

(a) The maximum potential difference across the resistor is 339.41 V.

(b) The maximum current through the resistor is 0.23 A.

(c) The rms current through the resistor is 0.16 A.

(d)  The average power dissipated by the resistor is 38.4 W.

<h3>Maximum potential difference</h3>

Vrms = 0.7071V₀

where;

  • V₀ is peak voltage

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

<h3> rms current through the resistor </h3>

I(rms) = V(rms)/R

I(rms) = (240)/(1,540)

I(rms) = 0.16 A

<h3>maximum current through the resistor </h3>

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

<h3> Average power dissipated by the resistor</h3>

P = I(rms) x V(rms)

P = 0.16 x 240

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Learn more about maximum current here: brainly.com/question/14562756

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