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Give expression for the force on a current carrying conductor in a magnetic field.
Solution
As shown in figure, consider a conductor $PQ$ of length $l$ are of cross-section $A$, carrying current $I$ along + ve $y$-direction. The field $\overrightarrow{\mathrm{B}}$ acts along + ve $z$-direction.
The electrons drift towards left with velocity $\overrightarrow{v_{d}}$.
Each electron experience force along + ve X-axis which is given by,
$\vec{f}=-e\left(\overrightarrow{v_{d}} \times \overrightarrow{\mathrm{B}}\right)$
If $n$ is the number of free electrons per unit volume, then total number of electrons in the conductor is,
$\mathrm{N}=n \times$ Volume $=n \mathrm{Al}$
Total force on the conductor is,
$\overrightarrow{\mathrm{F}}=\mathrm{N} \vec{f}=n \mathrm{Al}\left[-e\left(\overrightarrow{v_{d}} \times \overrightarrow{\mathrm{B}}\right)\right]$
$=n \mathrm{~A} e\left(-\left(l \overrightarrow{v_{d}} \times \overrightarrow{\mathrm{B}}\right)\right)$
But I $\vec{l}$ represents a current element vector in the direction of current so we can take
$\vec{v}_{d}=v_{d} \vec{l}$
$\therefore \quad\overrightarrow{\mathrm{F}}=n \mathrm{Ae}\left(v_{d} \vec{l} \times \overrightarrow{\mathrm{B}}\right)$
$=n \mathrm{Aev}_{d}(\vec{l} \times \overrightarrow{\mathrm{B}})$
but $n$ Aev $_{d}=$ current $I$
$\therefore \vec{F}=\mathrm{I}(\vec{l} \times \overrightarrow{\mathrm{B}})$
and magnitude $\mathrm{F}=\mathrm{I} l \mathrm{~B} \sin \theta$
where $\theta$ is angle between $\overrightarrow{\mathrm{B}}$ and $\mathrm{I}$.
This equation can be applicable for straight conducting rod.
If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips $d l$ and summing,
$\therefore \overrightarrow{\mathrm{F}}=\sum_{i=1}^{n} \overrightarrow{d l}_{i} \times \overrightarrow{\mathrm{B}}$ where, $i=1,2,3, \ldots, n$
Similar Questions
A charged particle (electron or proton) is introduced at the origin $(x=0, y=0, z=0)$ with a given initial velocity $\overrightarrow{\mathrm{v}}$. A uniform electric field $\overrightarrow{\mathrm{E}}$ and magnetic field $\vec{B}$ are given in columns $1,2$ and $3$ , respectively. The quantities $E_0, B_0$ are positive in magnitude.
|
column $I$ |
column $II$ | column $III$ |
| $(I)$ Electron with $\overrightarrow{\mathrm{v}}=2 \frac{\mathrm{E}_0}{\mathrm{~B}_0} \hat{\mathrm{x}}$ | $(i)$ $\overrightarrow{\mathrm{E}}=\mathrm{E}_0^2 \hat{\mathrm{Z}}$ | $(P)$ $\overrightarrow{\mathrm{B}}=-\mathrm{B}_0 \hat{\mathrm{x}}$ |
| $(II)$ Electron with $\overrightarrow{\mathrm{v}}=\frac{\mathrm{E}_0}{\mathrm{~B}_0} \hat{\mathrm{y}}$ | $(ii)$ $\overrightarrow{\mathrm{E}}=-\mathrm{E}_0 \hat{\mathrm{y}}$ | $(Q)$ $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{x}}$ |
| $(III)$ Proton with $\overrightarrow{\mathrm{v}}=0$ | $(iii)$ $\overrightarrow{\mathrm{E}}=-\mathrm{E}_0 \hat{\mathrm{x}}$ | $(R)$ $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{y}}$ |
| $(IV)$ Proton with $\overrightarrow{\mathrm{v}}=2 \frac{\mathrm{E}_0}{\mathrm{~B}_0} \hat{\mathrm{x}}$ | $(iv)$ $\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \hat{\mathrm{x}}$ | $(S)$ $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{z}}$ |
($1$) In which case will the particle move in a straight line with constant velocity?
$[A] (II) (iii) (S)$ $[B] (IV) (i) (S)$ $[C] (III) (ii) (R)$ $[D] (III) (iii) (P)$
($2$) In which case will the particle describe a helical path with axis along the positive $z$ direction?
$[A] (II) (ii) (R)$ $[B] (IV) (ii) (R)$ $[C] (IV) (i) (S)$ $[D] (III) (iii)(P)$
($3$) In which case would be particle move in a straight line along the negative direction of y-axis (i.e., more along $-\hat{y}$ )?
$[A] (IV) (ii) (S)$ $[B] (III) (ii) (P)$ $[C]$ (II) (iii) $(Q)$ $[D] (III) (ii) (R)$
