At what speed in ms will a proton move in a circular path of...

At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.95 × 10^6 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.25 × 10^-5 T? (b) What would the radius (in m) of the path be if the proton had the same speed as the electron? (c) What would the radius (in m) be if the proton had the same kinetic energy as the electron? (d) What would the radius (in m) be if the proton had the same momentum as the electron?

Transcript

00:01 Hello students, in this question we are utilizing these constants i .e.

00:03 Charge of proton, mass of proton, charge of electron and mass of electron.

00:08 And we are using the equation that is mv square by r i .e.

00:13 The centripetal force is equal to magnetic force q into v into b.

00:19 In this case, we have given velocity of electron is 7 .95 into 10 raise to 6 meter per second and also magnetic field intensity is given as 1 .25 into 10 raise to minus 5 tesla.

00:35 First we have to find the radius.

00:37 So r e is equal to mv square divided by q into v into b which is equal to mass of electron into velocity of electron divided by q e into b.

00:52 So now substituting the values provided with us i .e.

00:56 Mass v q and b, we have radius is equal to 3 .61 meter.

01:02 So in the first question, the same radius for the proton.

01:07 So here r p is equal to 3 .61 meter.

01:12 We have to find the velocity.

01:14 So v p is equal to q p into b into r p divided by mass of proton.

01:23 So we have to substitute the values and we get v p is equal to 4 .32 into 10 raise to 3 meter per second.

01:35 So first question is able to find the speed of the proton.

01:40 So we have speed of proton is equal to 4 .32 into 10 raise to 3 meter per second.

01:49 Now we are looking at the case of the radius if the proton has same that of electron.

01:55 So in the second question, we have provided that velocity of electron is equal to velocity of proton which is equal to 7 .95 into 10 raise to 6 meter per second.

02:09 So we have to calculate r p.

02:11 So r p is equal to m p v p divided by q p into b...

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