X x x x x x -> v x x x 9. A positively...

x x x x x x -> v x x x 9. A positively charged particle enters a region containing a uniform magnetic field of 0.100 T. The mass is accelerated by applying an electric potential of 50.0 Volts. In the magnetic field region the mass moves along a circular path of radius 10.0 cm. a) Determine the charge to mass ratio for this particle (a number). b) In an independent experiment the charge on the particle is measured as 1.00 ?C. What is the speed of the particle? c) Determine the mass of the particle. d) What is the direction and magnitude of the force on the particle as it enters the field in the figure shown above?

Transcript

00:01 Here in this given problem, this is a uniform magnetic field which is directed into the plane of paper.

00:09 Within this magnetic field, a positive charge is moving perpendicular to the magnetic field and towards right.

00:20 That is question number nine which we are solving here magnitude of this magnetic field that is 0 .100 tesla speed of this positive charge with which it is moving towards right that is 50 .0 volt and it constitute a circular path within this magnetic field whose radius is 10 .0 centimeter or we can say this is 0 .100 meter.

00:58 Okay, in the first part of the problem we have to find charge to mass ratio of this positive charge particle for which we will be using an expression for the radius of the circular path constituted by the charge particle moving perpendicular to the magnetic and that radius is given by mv by bq.

01:19 M is the mass of that charged particle, q is the charge over it.

01:23 Here, this mv that is also known as momentum, linear momentum.

01:28 And linear momentum in terms of kinetic energy of the particle that is given as square root of 2 times of mass, 2 times of product of mass with the kinetic energy gained by the charged particle divided by same bq.

01:42 Or finally we can say this is two times of product of mass with the kinetic energy and as this charge is accelerated through a potential this is v capital v for potential it is accelerated through a potential difference 50 volt so kinetic energy gained by it that may be given as the product of the charge with the potential difference divided by bq now we can write it like this 1 by b and q we shift it within this square root so it will become q square so canceling this q over here finally r is equal to 1 by p multiplied by square root of 2 m v by q square in both the sides r square is equal to 2 again here this is mistake this is not small v this is capital v for voltage for potential so this is 2 v by v square and we write it separately what that is m by q so charge to mass ratio q by m will be given by an expression 2 v by b square r square so this is 2 times of potential which is 50 volt divided by b square means 0 .100 square and radius 0 .100 meter square so this charge to mass ratio finally calculated to be equal to 1 .0 into 10 raised to the power 6 coulomb per kg and that is the answer for the first part of this given problem here.

03:46 Okay, now in the second part of the problem, we have to find the speed of the charge particle within this magnetic field provided the charge is 1 .00 microcoulomb.

04:01 So using an expression for kinetic energy either it is given as in mechanics this is simply half m v square and here this kinetic energy is actually the product of the magnitude of the charge and the potential difference.

04:18 So from here we get an expression for the speed of the charged particle and that is 2qv divided by m square root of 2qv by m or if we rearrange it we may write it like q by m charged to mass ratio multiplied by 2v...

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