00:01
Hello everyone we are going to understand this question here given in the question given a positively charged particle so mass of positively charged particle mass of positively charged particle that is represented by m which is 7 .2 into 10 to the power 8 kilogram 10 to 0 .m minus 8 kilogram and it is traveling due east with speed 80 meter 85 meter per second speed of charged particle that is represented by v which is 85 meter per second and it is traveling towards east and magnitude of uniform magnetic field magnitude of uniform magnetic field that is represented by b which is 0 .31 texla the particle moves through one quarter of a circle.
01:51
So time t is equal to 2 .2 into 10 to the power minus 3 second.
02:00
2 .2 into 10 to the 4 minus 3 second.
02:06
So at which time it leaves the field heading due south, all during motion, all during the motion of the particle moves perpendicular to the magnetic field.
02:24
So what is the magnitude of magnetic force acting on the particle? so magnitude of magnetic force f is equal to qv cross b.
02:44
Now substituting the value in this here value of q is not given.
02:59
So first of all we will calculate the charge on the particle.
03:02
So here it is moving on the curd path.
03:06
It enters into the magnetic field.
03:11
It enters into the magnetic field due east and out from the due south and it completes one quarter circle.
03:23
So we can write centrifugal force is equal to magnetic force.
03:32
It means qvv is equal to mv squared.
03:36
Square upon r.
03:38
So from this we can write r is equal to mv square.
03:45
Here v and we will get cancelled so we will get mv upon qv...