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CBSE Class 12 Physics 2023 Delhi Set 1 Paper

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Question : 23 of 35
Marks: +1, -0
(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 r2.
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.
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=µ0I2πd
The direction of field is
Force acting on the particle =q(v×B)
Here, θ=90
So, Force =qvB=µ02πqvdI
Its direction is towards right. Repulsive.
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