Describe the motion of a charged particle in a cyclotron if the frequency of the radio frequency (rf) field were doubled.
Here, the condition of magnetic resonance is violated.
When the frequency of the radio frequency (rf) field were doubled, the time period of the radio frequency ( rf ) field were halved. Therefore, the duration in which particle completes half revolution inside the dees, radio frequency completes the cycle.
Hence, particle will accelerate and decelerate alternatively. So, the radius of path in the dees will remain same.
Two long wires carrying current $I_1$ and $I_2$ are arranged as shown in figure. The one carrying current $I_1$ is along is the $x$-axis. The other carrying current $I_2$ is along a line parallel to the $y$-axis given by $x=0$ and $z=d$. Find the force exerted at $o_2$ because of the wire along the $x$-axis.
In Biot- Savart law, magnetic field $\mathbf{B}$ is parallel to $i \mathbf{d} \mathbf{l} \times \mathbf{r}$ and $i \mathbf{d} \mathbf{l}$ have its direction along the direction of flow of current.
Here, for the direction of magnetic field, At $\mathrm{O}_2$, due to wire carrying $I_1$ current is
$$\text { B } \| \text { parallel } i \text { dl } \times r \text { or } \hat{\mathbf{i}} \times \hat{\mathbf{k}} \text {, but } \hat{\mathbf{i}} \times \hat{\mathbf{k}}=-\hat{\mathbf{j}}$$
So, the direction at $\mathrm{O}_2$ is along $Y$ - direction.
The direction of magnetic force exerted at $\mathrm{O}_2$ because of the wire along the, $x$-axis.
$$\mathbf{F}=I l \times \mathbf{B} \approx \hat{\mathbf{j}} \times(-\hat{\mathbf{j}})=0$$
So, the magnetic field due to $l_1$ is along the $y$-axis. The second wire is along the $y$-axis and hence, the force is zero.
A current carrying loop consists of 3 identical quarter circles of radius $R$, lying in the positive quadrants of the $x-y, y-z$ and $z-x$ planes with their centres at the origin, joined together. Find the direction and magnitude of $\boldsymbol{B}$ at the origin.
For the current carrying loop quarter circles of radius $R$, lying in the positive quadrants of the $x$-y plane
$$\mathrm{B}_1=\frac{\propto_0}{4 \pi} \frac{I(\pi / 2)}{R} \hat{\mathbf{k}}=\frac{\propto_0}{4} \frac{I}{2 R} \hat{\mathbf{k}}$$
For the current carrying loop quarter circles of radius $R$, lying in the positive quadrants of the $y$-z plane
$$\mathrm{B}_2=\frac{\propto_0}{4} \frac{I}{2 R} \hat{\mathrm{i}}$$
For the current carrying loop quarter circles of radius $R$, lying in the positive quadrants of the z-x plane
$$\mathrm{B}_3=\frac{\propto_0}{4} \frac{I}{2 R} \hat{\mathrm{i}}$$
Current carrying loop consists of 3 identical quarter circles of radius $R$, lying in the positive quadrants of the $x-y, y-y$ and $z-z$ planes with their centres at the origin, joined together is equal to the vector sum of magnetic field due to each quarter and given by
$$\mathbf{B}=\frac{1}{4 \pi}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \frac{\propto_0 I}{2 R} .$$
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.