CBSE 2023 · Region 5 · Set 2 · Q23 · 2 marks
Write the expression for the Lorentz force on a particle of charge q moving with a velocity $\displaystyle \vec{v}$ in a magnetic field $\displaystyle \vec{B}$. 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 $\displaystyle \vec{r}_{1}$ to point $\displaystyle \vec{r}_{2}$.A long straight wire AB carries a current I. A particle (mass m and charge q) moves with a velocity $\displaystyle \vec{v}$, 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. (Figure: a long vertical wire from A at the bottom to B at the top carrying current I upwards; a particle sits a horizontal distance d to the right of the wire, moving with velocity v directed downwards, parallel to the wire.)
Write the expression for the Lorentz force on a particle of charge q moving with a velocity $\displaystyle \vec{v}$ in a magnetic field $\displaystyle \vec{B}$. 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 $\displaystyle \vec{r}_{1}$ to point $\displaystyle \vec{r}_{2}$.
A long straight wire AB carries a current I. A particle (mass m and charge q) moves with a velocity $\displaystyle \vec{v}$, 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. (Figure: a long vertical wire from A at the bottom to B at the top carrying current I upwards; a particle sits a horizontal distance d to the right of the wire, moving with velocity v directed downwards, parallel to the wire.)
Marking-scheme solution
(a)
\[\vec{F}_m = q\left(\vec{v} \times \vec{B}\right)\]
\[F_m = q\,v\,B \sin\theta \qquad (\because\ \vec{v} \perp \vec{B})\]
Force is maximum for $\displaystyle \theta = 90^{0}$
As magnetic force always acts in a direction perpendicular to the velocity vector, hence no work is done by this force on the particle during motion.
Alternatively $\displaystyle W = \vec{F}\cdot(\vec{r_2} - \vec{r_1})$
\[= F\left|\vec{r_2} - \vec{r_1}\right|\cos 90^{\circ}\]
\[= 0\]
OR
(b) The magnetic field produced by current carrying conductor AB\[B = \frac{\mu_0 I}{2\pi d}\] , directed into the plane of paper.
Force experienced by charged particle
\[\vec{F_m} = q\left(\vec{v} \times \vec{B}\right)\]
\[\therefore\ \vec{v} \perp \vec{B}\]
\[F_m = q\,v B \sin 90^{0}\]
\[F_m = q\,v \times \frac{\mu_0 I}{2\pi d}\]
Force is repulsive/acts towards right / away from the conductor.
Moving Charges and MagnetismMagnetic ForceApplyvery_short_answermedium
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CBSE Class 12 Physics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.