Physics, 07.05.2020 05:57 Anybody6153

A cross section of three parallel wires each carrying a current of 20 A. The current in wire B is out of the paper, while that in wires A and C are into the paper. If the distance R = 15.0 mm,

what is the magnitude of the force on a 200 cm length of wire C?

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Answer from: Reganlist

Complete Question

The diagram for this question is shown on the first uploaded image

The magnitude of force is $F_R} = 5.33 \ m N$

Explanation:

From the question we are told that

The current in wire A , B and C are I _a = I_b =I_c = I= 20 A

The distance is $R = 15.0mm = \frac{15}{1000} = 0.015m$

The length of wire C is $L_c = 200cm = \frac{200}{100} = 2 m$

Generally the force exerted per unit length that is acting in between two current carrying conductors can be mathematically represented as

$\frac{F}{L} = \frac{\mu_o }{2 \pi} \cdot \frac{I_1 I_2 }{R}$

Where $\mu_o$ is the permeability of free space with a constant value of

$\mu_o = 4 \pi * 10^{-7} N/A^2$

When the current in the wire are of the same direction the force is positive and when they are in opposite direction the force is negative

Considering force between A and C

$F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L$

Considering force between B and C

$F_{B -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L$

The resultant force is

$F_R = F_{B -C} - F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L - \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L$

$F_R} = \frac{\mu_o }{2 \pi} \cdot \frac{I * I }{R} * L * ({1 - \frac{1}{2} })$

Substituting values

$F_R} = \frac{4 \pi * 10^{-7}}{2 * 3.142} \cdot \frac{ 20*20 }{0.015} * 2 ({\frac{1}{2} })$

$F_R} = 5.33 *10^{-3} N$

$F_R} = 5.33 \ m N$

A cross section of three parallel wires each carrying a current of 20 A. The current in wire B is ou

Answer from: Quest

Iread this book last year so take this with a grain of salt.gene jumps from the tree to compete with finny due to the fact that gene is always trying to do better than finny. he is put under the grasp of peer pressure due to finny almost forcing him to do this. ever since he first jumped finny kept pressuring him to jump since he thought gene enjoyed jumping.rather than uses this answer you should probably just reread chapters 3-4

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$