A uniform beam of positively charged particles is moving with a constant velocity parallel to another beam of negatively charged particles moving with the same velocity in opposite direction separated by a distance $d.$ The variation of magnetic field $B$ along a perpendicular line draw between the two beams is best represented by

$\alpha $ particle, proton and duetron enters in a uniform (transverse) magnetic field $'B'$ with same acceleration potential find ratio of radius of path followed by these particles.

If a proton is projected in a direction perpendicular to a uniform magnetic field with velocity $v$ and an electron is projected along the lines of force, what will happen to proton and electron

A beam of well collimated cathode rays travelling with a speed of $5 \times {10^6}\,m{s^{ - 1}}$ enter a region of mutually perpendicular electric and magnetic fields and emerge undeviated from this region. If $| B |=0.02\; T$, the magnitude of the electric field is

A particle of charge $q$ and velocity $v$ passes undeflected through a space with non-zero electric field $E$ and magnetic field $B$. The undeflecting conditions will hold if.

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 $\vec{B}$ ?