A charged particle is projected in magnetic field B=(3i+4j) x 10-2 T accelarationof particle is found to be a=xi+2j - Physics - Magnetism And Matter

Question

A charged particle is projected in magnetic field B=(3i+4j) x
10- 2 T accelarationof particle is found to
be a=xi+2j m/s2 Find x

Ved Prakash Lakhera · Accepted Answer

magnetic field B = 3i^+4j^×10-2T   and the acceleleration a = xi^+2j^so when charged particle moves in this magnetic field then,               mv2r=Bqv, where r is the fixed radius along which charged particle moves
Therefore        r = mvBq, on differentiating aboe eqn with respect to time, we get,                       drdt = mq1B×dvdt=mqaB
But dr/dt =0 , so  aB=0, Therefore     xi^+2j^3i^+4j^×10-2=0or                xi^+2j^  
 3i^+4j^-1×102 =0or                    xi^+2j^
3i^-4j^=0or                   3x - 8=0or                    x = -8/3

A charged particle is projected in magnetic field B=(3i+4j) x 10- 2 T. accelarationof particle is found to be a=xi+2j m/s2. Find x.

A charged particle is projected in magnetic field B=(3i+4j) x 10^- ²T. accelarationof particle is found to be a=xi+2j m/s². Find x.