Three very long parallel wires a small portion of each is...

Three very long, parallel wires (a small portion of each is shown in the drawing) are resting on a flat surface. The distance between wire B, which has a 15 mA current to the left, and its neighbors is 0.0015 m. Wire A carries a 10 mA current toward the right; and wire C carries a 5 mA current toward the right. Rank the wires in order of the magnitude of the net magnetic force on each, with the largest value first and the lowest value last. A 10 mA B 15 mA C 5 mA Lütfen birini seçin: a. A>C> B b. B>A>C c. A>B>C d. C>B>A e. B>C>A

Transcript

00:01 Hi, this given problem is based upon the force experienced by the two parallel current carrying conductors.

00:08 And that force, this is given by the expression mu not upon 4 pi into 2 i1 into i2 by r.

00:16 R is the gap between them.

00:17 And one more concept which we will use here, the conductors which are carrying currents in the same direction.

00:42 They attract each other and those carrying current in opposite directions they repel each other so here looking at the first conductor conductor a the direction of current in a and b these are in opposite direction so they will repel each other so force experienced by a due to conductor b f at a due to b that is in opposite direction means going up and the force experienced by a due to c as the current are in the same direction so that will be an attractive force force experienced by a due to c but the magnitude of facc will be less than fap because the force depends inversely on the gap between the two conductors now taking wire b it will be repelled by conductor a by a force f at b due to a, due to a, and it will be repelled by the conductor c also, f at b due to c.

02:09 Now, the third conductor c, it will be repelled by conductor b, f at c due to b.

02:17 It will be attracted by conductor a, f at c due to a.

02:22 So, the forces experienced by all the three conductors, these are in opposite direction.

02:27 We just need to find their magnitudes.

02:30 So starting with the force experienced by conductor afa, it will be given by fab minus fac.

02:42 Now using the expression and in it taking this new not upon 4 pi and 2 as a common out, leaving behind i1, i2 by r.

02:53 So here it will become 10 multiplied by 15.

02:58 And the currents are in millie amper.

03:01 So 10 to the bar minus 9 will be taken out as a common.

03:06 And divided by r, the distance between them, this is given as 0 .0015 minus fac.

03:18 For it, the currents are 10 and 5, 10 multiplied by 5 divided by gap between them...

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