A very long, thin rod carries electric charge with the...

A very long, thin rod carries electric charge with the linear density 42.0 nC/m. It lies along the x axis and moves in the x direction at a speed of 1.50 !!! 10^7 m/s. (a) Find the electric field the rod creates at the point (x = 0, y = 22.0 cm, z = 0.) magnitude N/C direction (b) Find the magnetic field it creates at the same point. magnitude T direction (c) Find the force exerted on an electron at this point, moving with a velocity of (2.40 !!! 10^8)i m/s. magnitude N direction

Transcript

00:01 Linear charge density that is 42 nanoculum per meter it can be written as 42 into 10 raise to power minus 9 coulum per meter and the velocity is given to us 1 .5 into 10 raised to power 7 meter per second as it is a thin rod so let's say length will be 1 meter and now if we have to find the electric filter at point where x equals to 0 y is equal to 22 centimeter and then z equals to 0 so that can be written as 0 .22 meter now further using we know that phi e .da which is also equals to q enclosed by epsilon node area for rod can be treated as electric field into 2 pi r which is equals to lambda l which is q and then epsilon not now further shifting towards right hand side that is lambda l by 2 pi epsilon not are so putting up given values 42 into 10 raise to power minus 9 1 then 2 into 3 .14 into 8 .85 into 10 raise to power minus 12 into 0 .22 as 3 .43 into 10 raise to power 3 j cap newton per column and that is the direction and that is the required answer.

01:53 Now further for magnetic field at same point current will be which is lambda into velocity that is linear charge density 42 into 10 raise to power minus 9 coulum per meter and then velocity which is 1 .5 into 10 days to power 7 meter per second so meter is getting cancelled then current will be that is 0 .63 amp now magnetic field is mu not i upon 2 pi r so putting up values that is mu not upon 4 pi you can also treat it as 2 mu not i upon 4 pi r then mu not upon 4 pi is that is 2 which is into 10 raised to power minus 7 and then i is 0 .63 divided by that is r is 0 .22 so further upon solving 0 .7 into 10 raised to power minus 7 tesla that is the required magnetic field of electro.

03:13 With velocity 2 .4 into 10 raised to power 8 meter per second.

03:20 That comes out to be combined force qe plus qv cross b...

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