Q.

A positively charged particle P crosses a region of uniform electric field of intensity E. Velocity u of the particle at the entry on the left side of the region is perpendicular to the electric field as shown in the first figure. Speed of the particle at exit on the right side is v1 . Now the experiment is repeated with a uniform magnetic field of induction B superimposed on the electric field as shown in the second figure, keeping entry velocity of the particle unchanged. In this experiment, speed of the particle at exit on the right side is v2 . What can you conclude regarding the speeds v1  and v2 ?
 

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a

If u=EB,  then  v2<v1

b

If u<EB,  then v2<v1

c

If u>EB,  then  v2<v1

d

If  u=EB, then v2=v1  

answer is A, B, D.

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Detailed Solution

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If both E  and B  are present:
Case I:  u=EBFnet=0v2=u<v1 
Case II:   u>EBFm>FE
Particle turns in the direction opposite to E  and electric field does negative work. 

v2<u<v1

Case III:   u<EBFe>Fm
Particle turns in the direction of E  and electric field does positive work. But displacement in the direction of  E will be less than case of only E.
v1>v2>u
In all three cases,  v2<v1

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