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CBSE Class 12 Physics 2023 Delhi Set 1 Paper

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Question : 23
Total: 35
(a) Write the expression for the Lorentz force on a particle of charge q moving with a velocity
→
v
 in a magnetic field
→
B
. When is the magnitude of this force maximum? Show that no work is done by this force on the particle during its motion from a point
→
r1
 to point
→
r
2.
OR
(b) A long straight wire AB carries a current I. A particle (mass m and charge q ) moves with a velocity
→
v
, parallel to the wire, at a distance d from it as shown in the figure. Obtain the expression for the force experienced by the particle and mention its directions.
Solution:     👈: Video Solution
Expression for Lorentz force:
→
F
=q(
→
v
×
→
B
)
The force is maximum when the angle between
→
v
 and
→
B
 is 90∘.
Here,
→
F
 is perpendicular to
→
v
. So, no work is done by this force on the particle during its motion
OR
(b) Magnetic field produced by the current carrying from a point
→
r1
 to point
→
r2
 wire, B=‌
µ0I
2Ï€d

The direction of field is ⊗
Force acting on the particle =q(
→
v
×
→
B
)
Here, θ=90∘
So, ‌ Force ‌=qvB=‌
µ0
2Ï€
‌
qv
d
I
Its direction is towards right. Repulsive.
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