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ryzh [129]
2 years ago
8

A solenoid has a length , a radius , and turns. The solenoid has a net resistance . A circular loop with radius is placed around

the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance . At a time , the solenoid is connected to a battery that supplies a potential . At a time , what current flows through the outer loop
Physics
1 answer:
geniusboy [140]2 years ago
7 0

This question is incomplete, the complete question is;

A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?

Answer:

the current flows through the outer loop is 1.3 × 10⁻⁵ A

Explanation:

Given the data in the question;

Length l = 11.34 cm = 0.1134 m

radius a = 1.85 cm = 0.0185 m

turns N = 1627

Net resistance R_{sol = 144.9 Ω

radius b = 3.77 cm = 0.0377 m

R_o = 1651.6 Ω

ε = 34.95 V

t = 2.58 μs = 2.58 × 10⁻⁶ s

Now, Inductance; L = μ₀N²πa² / l

so

L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134

L = 0.003576665 / 0.1134

L = 0.03154

Now,

ε = d∅/dt = \frac{d}{dt}( BA ) =  \frac{d}{dt}[ (μ₀In)πa² ]

so

ε = μ₀n \frac{dI}{dt}( πa² )

ε = [ μ₀Nπa² / l ] \frac{dI}{dt}

ε = [ μ₀Nπa² / l ] [ (ε/L)e^( -t/R_{sol) ]

I = ε/R_o = [ μ₀Nπa² / R_ol ] [ (ε/L)e^( -t/R_{sol) ]

so we substitute in our values;

I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]

I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]

I = 1.3 × 10⁻⁵ A

Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A

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