8. In a linear accelerator, protons are accelerated from...

8. In a linear accelerator, protons are accelerated from rest through a potential difference to a speed of approximately 1.5×10^5 m/s. The proton beam enters a region of uniform magnetic field B, as shown below, that causes the beam to follow a semicircular path. Determine the magnitude of the field that is required to cause an arc of radius 0.3 m. The charge of the proton is 1.60 × 10^-19C, and the mass of the proton is 1.67×10^-27 kg.

Transcript

00:01 Here in this question a linear accelerator is given in which the protons are accelerated.

00:08 So here the charge used this proton in the linear accelerator.

00:16 So here the protons are moving with a speed of, that is speed is given as 1 .5 into 10 power 5 meter per second.

00:29 Also the mass of the proton is given.

00:31 Mass of the proton is 1 .67 into 10 power minus 27 kilogram which is a constant value also the charge of proton is given the charge of proton q is equal to q represents the charge of proton which is given as 1 .60 into 10 power minus 19 kulumb the si unit of charge is kulumb similarly the proton upon entering a region of uniform magnetic field is it follows a semi -circular path and the radius of the path is given as the radius of the path is given as 0 .3 meter.

01:15 So this is how it is actually sorry the proton is moving.

01:20 So here there is a region of magnetic field which is yellow in color.

01:23 So on reaching that magnetic field the proton moves in a semi -circular path.

01:28 This is the semi -circular path and the radius is given.

01:31 So if this is the case, then they are asking to calculate the magnitude of the magnetic field.

01:38 That is magnitude of the magnetic field is to be calculated here.

01:43 So here we know that here the proton is following a semicircular path.

01:49 So definitely there will be a net centripetal force...