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Describe the motion of a charged particle in a combined electric and magnetic field. |
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Answer» Solution :Let E be the uniform electric field applied in the y direction and B be the uniform magnetic FILED applied along the z direction. Let the change q MOVE with a speed v along the X direction. i.e., `VEC(E)=Ehat(j), vec(B)=Bvec(k) " and " vec(v)=vhat(i)` `therefore ""vec(F)_(e)=qEvec(j)` (electric force) magnetic force `""vec(F)_(mag)=qvec(v) times B=qvhat(i) times Bhat(k)` i.e.,`""vec(F)_(m)=qvB(-vec(j))` By applying Fleming's Left Hand Rule, we note that the resultant force on the charged particle `""vec(F)=qEhat(j)-qvBhat(j)` i.e.,`""vec(F)=q(E-vB)hat(j)` If QE = qvB, then net force on the change will be zero and the charge continues to move along x direction without deflection. Since qE = qvB, `v=E/B` where 'v' is called the velocity. E and B are called crossed fields. The crossed fields `vec(E) " and " vec(B)` may be used to select charged particles of a particular velocity out of a beam containing charges moving with different speeds. 
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