Question

A charged particle (-q, m) is projected with velocity of Ū in direction of uniform electric field of strength E in horizontal plane. Time in which velocity of particle becomes - Ū is ?​

Answer 1

Explanation:

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Answer 2

Answer:

Correct option is

A

2E

BE

is an integer

Charge =q

m

Velocity =u

i

^

−u

j

^

along the +y direction E and B.

The particle will definitely return to the origin once if.

In Y direction the electric field E exist. So, the force due to electric on the charge is given by

F=qE

Now, the acceleration of the charge is given by a=

m

qE

now, the acceleration on is opposite to the direction of velocity. So, it will return to origin if displacement in y− direction is zero.

d=V

y

×t+

2

1

at

2

0=−V×t+

2

1

m

qE

T

2

T=

qE

2mV

now due to magnitude field it will move in circular path, time period of its circular motion is given by

T=

qB

2πm

now if complete N number of rounds in above time of return then

N×

qB

2πm

=

qE

2mV

N=

πE

Bv

So, here above value must be an integer so that it complete integral number of rounds

integer =

πE

Bv

=

2E

Bv

(∵ n=2) in an integer.

Explanation:

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