Test Index

CBSE Class 12 Physics 2023 Delhi Set 1 Paper

© examsnet.com
Question : 23 of 35
Marks: +1, -0
(a) Write the expression for the Lorentz force on a particle of charge qq moving with a velocity v⃗\vec{v} in a magnetic field B⃗\vec{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 r⃗1\vec{r}_1 to point r⃗2\vec{r}_2.
OR
(b) A long straight wire ABAB carries a current I. A particle (mass mm and charge qq ) moves with a velocity v⃗\vec{v}, parallel to the wire, at a distance dd 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:
F⃗=q(v⃗×B⃗)\vec{F}=q(\vec{v}\times \vec{B})
The force is maximum when the angle between v⃗\vec{v} and B⃗\vec{B} is 90∘90^{\circ}.
Here, F⃗\vec{F} is perpendicular to v⃗\vec{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 r⃗1\vec{r}_1 to point r⃗2\vec{r}_2 wire, B=μ0I2πdB=\frac{\mu_0 I}{2\pi d}
The direction of field is ⊗\otimes
Force acting on the particle =q(v⃗×B⃗)=q(\vec{v}\times \vec{B})
Here, θ=90∘\theta=90^{\circ}
So,    Force   =qvB=μ02π  qvdI\; \text{ Force } \; = q v B = \frac{\mu_0}{2\pi} \; \frac{q v}{d} I
Its direction is towards right. Repulsive.
© examsnet.com
Go to Question:
Prev Question
Next Question