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A particle of specific charge a(charge per unit mass) is released from origin at t=0 with an initial velocity of v=v0i (vector) in a uniform magnetic field b=b0k(vector).Find velocity and position of particle at any time`t`.physics

Harshit Singh , 7 Years ago
Grade 12
anser 1 Answers
Parth

Last Activity: 7 Years ago

So, we have: q/m=a Velocity= v0i Field= b0kForce= q(V x B)So force would be -(v0b0)j Which is negative y direction.So, FOR POSITION:The motion would be on xy plane and a circular motion.Let`s consider a circle with centre at origin and radius R and let it make angle ø with the horizontal.so coordinates are :x=Rcosø and y= -Rsinø(as motion is on negative y direction)And ø= wt(w is angular speed of particle)And w= (bq)/m = ba(a is specific charge)Now as we have assumed centre as origin but it is to be shifted R distance towards negative y axis.So the coordinates are:X=Rcos(wt), Y= -(R+Rsin(wt))Now,FOR VELOCITY:AS FORCE IS PERPENDICULAR MAGNITUDE OF VELOCITY REMAINS UNCHANGED.NOW THE DIRECTION OF VELOCITY WOULD BE PERPENDICULAR TO RADIUS VECTOR .RADIUS UNIT VECTOR= (cos wt i , -sinwt j )Unit VECTOR PERPENDICULAR to radius VECTOR would be:-sinwt i -coswt j So velocity is-v0 (sin wt i cost j)

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