moving-charges-and-magnetism Question 25

Question: Q. 1. Find the condition under which the charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed.

R [O.D. I, II, III 2017]

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Solution:

Ans. (i) For directions of E,B,v

(ii) For magnitudes of E,B,v

(i) The velocity v of the charged particles, and the E and B vectors, should be mutually perpendicular. 1/2 Also the forces on q, due to E and B must be oppositely directed. (Also accept if the student draws a diagram to show the directions.)

(ii) qE=qvB

1/2

v=EB

1/2

[Alternatively, The student may write :

Force due to electric field =qE

Force due to magnetic field =q(v×B)

The required condition is

qE=q(v×B) [ or E=(v×B)=(B×v)]

(Note : Award 1 mark only if the student just writes :

“The forces, on the charged particle, due to the electric and magnetic fields, must be equal and opposite to each other”)

[CBSE Marking Scheme 2017]

Detailed Answer :

If a charged particle travels in crossed electric and magnetic field, then both fields are perpendicular to each other.

Force on the particle will be :

F=FE+FB F=qE+q(v×B)

As FB is 180 to FE, so

F=qEq(v×B) F=q(EvvB)j^

Now to select a particle, making resultant force as zero, hence

Now,

0=q(EvB)j^

Hence,

EvB=0 E=vB

Hence,

v=EB



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