An electron is moving in the vicinity of a long, straight...

An electron is moving in the vicinity of a long, straight wire that lies along the x-axis. The wire has a constant current of 8.10 A in the x-direction. At an instant when the electron is at point (0, 0.200 m, 0) and the electron 's velocity is v = (5.00 imes 10^{4} ext{ m/s})hat{i} - (3.00 imes 10^{4} ext{ m/s})hat{j}. Part A What is the force that the wire exerts on the electron? Enter the x, y, and z components of the force separated by commas. Fx, Fy, Fz = Part B Calculate the magnitude of this force. Express your answer with the appropriate units. F =

Transcript

00:01 Hello everyone we are going to understand this question here given in the question given current in the wire current in wire that is represented by i which is 8 .10 ampere and position of electron that is 0 .0 .200 comma 0 .0 and electrons velocity that is represented by v, which is 5 .0 into 10 to the power 4 icap minus 3 .0 into 10 to the power 4 jcap.

01:22 Now in the part a, first of our we will calculate the magnetic field due to current carrying conductor.

01:32 So magnetic field due to current carrying conductor, magnetic field due to current carrying conductor, current carrying conductor that is represented by b, which is mu not i upon 2 pi into d.

02:09 Now multiplying and dividing by 2.

02:11 So 2 into mu not upon 4 pi into i upon d.

02:19 Now substituting the value in this.

02:22 So we can write 2 into 10 to the power minus 7 into value of i that is 8 .1 upon value of d that is 0 .2.

02:36 And direction of magnetic field is along to k direction.

02:42 So after doing further calculation we will get.

02:46 T1 into 10 to the power minus 7 tesla kkr.

02:51 Now we know that magnetic force f is equal to q into v cross v.

02:58 Magnetic force f is equal to q into v cross b.

03:10 Now substituting the value in this charge on electron that is 1 .6 into 10 to the power minus 19 into value of v that is 5 .0 into 10 to the power 4 i cap minus 3 .0 into 10 to the power 4 j cap into 801 into 10 to the power minus 7 k cap now doing further calculation so here we can write calculating v cross b so we will get we will get ijk value of i is 5 .0 into value of v 5 .0 into 10 4 and value of for v that is minus 3 .0 into 10 to the power 4 and value of k is 0 for velocity and for magnetic field value of i 0 0 and this is 81 into 10 to the power minus 7 10 to the power minus 7.

04:40 Let's correct it.

04:43 It's 10 to the power minus 7.

04:47 Now opening this determinants.

04:51 So here, i into minus 81 into 3 into 10 to the power 3.

05:04 I cap plus j into 5 .0 into 81 into 10 to the power 3.

05:15 And k term is 0 so after doing further calculation we will get 243 into 10 to the power minus 3 i cap plus 405 into 10 to the power minus 3 j now doing further calculation hence the required value of force along to x axis hence the required value of force required value of force along to x -axis that is represented by f x which is 1 .6 into 10 to the power minus 19 into 243 into 10 to the power minus 3 sorry this is minus 3 now doing further calculation so we will get f x is equal to so after calculation we will get this is minus and this one is also minus that is after calculation we got this value so after multiplying we will get 388 .8 .388 into 10 to the power minus 3 i cap this is the force along to x -axis and magnetic force along to y -axis that is 1 .6 into 10 to the power minus 19 into 405 sorry, here we forgot to write this is 10 to the power minus 19 plus 21.

07:26 That will give 10 to the power minus 19 minus 3...

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