An electron moving with a uniform velocity along the positive $x$-direction enters a magnetic field directed along the positive $y$-direction. The force on the electron is directed along

A proton is projected with velocity $\overrightarrow{ V }=2 \hat{ i }$ in a region where magnetic field $\overrightarrow{ B }=(\hat{i}+3 \hat{j}+4 \hat{k})\; \mu T$ and electric field $\overrightarrow{ E }=10 \hat{ i } \;\mu V / m .$ Then find out the net acceleration of proton (in $m / s ^{2}$)

Two parallel wires in the plane of the paper are distance $X _0$ apart. A point charge is moving with speed $u$ between the wires in the same plane at a distance $X_1$ from one of the wires. When the wires carry current of magnitude $I$ in the same direction, the radius of curvature of the path of the point charge is $R_1$. In contrast, if the currents $I$ in the two wires have direction opposite to each other, the radius of curvature of the path is $R_2$.

If $\frac{x_0}{x_1}=3$, the value of $\frac{R_1}{R_2}$ is.

In the product

$\overrightarrow{\mathrm{F}} =\mathrm{q}(\vec{v} \times \overrightarrow{\mathrm{B}})$

$=\mathrm{q} \vec{v} \times\left(\mathrm{B} \hat{i}+\mathrm{B} \hat{j}+\mathrm{B}_{0} \hat{k}\right)$

For $\mathrm{q}=1$ and $\vec{v}=2 \hat{i}+4 \hat{j}+6 \hat{k}$ and

$\overrightarrow{\mathrm{F}}=4 \hat{i}-20 \hat{j}+12 \hat{k}$

What will be the complete expression for $\vec{B}$ ?

Fill the blank :

$(i)$ Static charge produces ...... field around it.(Electric, Magnetic)

$(ii)$ Moving charge produces ...... field around it.

An electron (mass = $9.1 \times {10^{ - 31}}$ $kg$; charge = $1.6 \times {10^{ - 19}}$ $C$) experiences no deflection if subjected to an electric field of $3.2 \times {10^5}$ $V/m$, and a magnetic fields of $2.0 \times {10^{ - 3}} \,Wb/m^2$. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius.......$m$