A charged particle of charge $e$ and mass $m$ is moving in an electric field $\mathbf{E}$ and magnetic field B. Construct dimensionless quantities and quantities of dimension $[T]^{-1}$.
No dimensionless quantity can be constructed using given quantities. For a charge particle moving perpendicular to the magnetic field, the magnetic Lorentz forces provides necessary centripetal force for revolution.
$$\frac{m v^2}{R}=q v B$$
On simplifying the terms, we have
$$\therefore \quad \frac{q B}{m}=\frac{v}{R}=\omega$$
Finding the dimensional formula of angular frequency
$$\therefore \quad[\omega]=\left[\frac{q B}{m}\right]=\left[\frac{v}{R}\right]=[T]^{-1}$$
This is the required expression.
An electron enters with a velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$ into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in plane parallel to the $x$-y plane. Suggest a configuration of fields $\mathbf{E}$ and $\mathbf{B}$ that can lead to it.
Considering magnetic field $\mathbf{B}=B_0 \hat{\mathbf{k}}$, and an electron enters with a velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$ into a cubical region (faces parallel to coordinate planes).
The force on electron, using magnetic Lorentz force, is given by
$$F=-e\left(v_0 \hat{i} \times B_0 \hat{k}\right)=e v_0 B_0 \hat{i}$$
which revolves the electron in $x$-y plane.
The electric force $\mathbf{F}=-\mathbf{e E _ { 0 }} \hat{\mathbf{k}}$ acceleratese along z-axis which in turn increases the radius of circular path and hence particle traversed on spiral path.
Do magnetic forces obey Newton's third law. Verify for two current elements $\mathbf{d l}_1=\mathbf{d l} \hat{\mathbf{i}}$ located at the origin and $\mathbf{d l}_2=\mathbf{d l} \hat{\mathbf{j}}$ located at $(0, R, 0)$. Both carry current $I$.
In Biot-Savart's law, magnetic field $\mathbf{B}$ is parallel (II) to $i \mathbf{d l} \times \mathbf{r}$ and $i \mathrm{dl}$ have its direction along the direction of flow of current.
$B \| i \mathbf{d l} \times \mathbf{r}$ or $\hat{\mathbf{i}} \times \hat{\mathbf{j}}$ (because point $(0, R, 0)$ lies on $y$-axis), but $\hat{\mathbf{i}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}}$
So, the direction of magnetic field at $d_2$ is along $z$-direction.
The direction of magnetic force exerted at $d_2$ because of the first wire along the $x$-axis. $F=i(I \times B) i . e ., F \|(i \times k)$ or along $-\hat{\mathbf{j}}$ direction.
Therefore, force due to $\mathrm{dl}_1$ on $\mathrm{dl}_2$ is non-zero.
Now, for the direction of magnetic field, At $d_1$, located at $(0,0,0)$ due to wire $d_2$ is given by $B \| i \mathbf{d l} \times \mathbf{r}$ or $\hat{\mathbf{j}} \times-\hat{\mathbf{j}}$ (because origin lies on $y$-direction w.r.t. point $(0, R, 0$ ).), but $\mathbf{j} \times-\mathbf{j}=0$.
So, the magnetic field at $d_1$ does not exist.
Force due to $\mathrm{dl}_2$ on $\mathrm{dl}_1$ is zero.
So, magnetic forces do not obey Newton's third law.
A multirange voltmeter can be constructed by using a galvanometer circuit as shown in figure. We want to construct a voltmeter that can measure $2 \mathrm{~V}, 20 \mathrm{~V}$ and 200 V using a galvanometer of resistance $10 \Omega$ and that produces maximum deflection for current of 1 mA . Find $R_1, R_2$ and $R_3$ that have to be used.
Applying expression in different situations
For $i_G\left(G+R_1\right)=2$ for 2 V range
For $i_G\left(G+R_1+R_2\right)=20$ for 20 V range
and For $i_G\left(G+R_1+R_2+R_3\right)=200$ for 200 V range
On solving, we get $R_1=1990 \Omega, R_2=18 \mathrm{k} \Omega$ and $R_3=180 \mathrm{k} \Omega$.
A long straight wire carrying current of 25 A rests on a table as shown in figure. Another wire $P Q$ of length 1 m , mass 2.5 g carries the same current but in the opposite direction. The wire $P Q$ is free to slide up and down. To what height will $P Q$ rise?
The magnetic field produced by long straight wire carrying current of 25 A rests on a table on small wire
$$B=\frac{\propto_0 I}{2 \pi h}$$
The magnetic force on small conductor is
$$F=B I l \sin \theta=B I l$$
Force applied on $P Q$ balance the weight of small current carrying wire.
$$\begin{aligned} & F=m g=\frac{\propto_0 I^2 l}{2 \pi h} \\ & h=\frac{\propto_0 I^2 l}{2 \pi m g}=\frac{4 \pi \times 10^{-7} \times 25 \times 25 \times 1}{2 \pi \times 2.5 \times 10^{-3} \times 9.8}=51 \times 10^{-4} \\ & h=0.51 \mathrm{~cm} \end{aligned}$$