(a) Write the expression for the Lorentz force on a particle of charge q moving with a velocity \overrightarrow{v} in a magnetic field \overrightarrow{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 \overrightarrow{r_1} to point \overrightarrow{r}_2. OR (b) A long straight wire {\AB} carries a current I. A particle (mass m and charge q ) moves with a velocity \overrightarrow{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.

CBSE Class 12 Physics 2023 Delhi Set 1 Paper - Question 23

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

{r_1}↖{→}

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:

Expression for Lorentz force:

{ {\F}}↖{→}=q({v}↖{→} × {B}↖{→})

The force is maximum when the angle between

{v}↖{→}

and

{B}↖{→}

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

{r}_1↖{→}

to point

{r}_2↖{→}

wire,

B=\;{µ_0 I}/{2 π d}

The direction of field is

⊗

Force acting on the particle

=q({v}↖{→} × {B}↖{→})

Here,

θ=90^{∘}

So,

\;\text " Force "\;=q v B=\;{µ_0}/{2 π} \;{q v}/{d} {\I}

Its direction is towards right. Repulsive.

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