$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ along the direction shown in the figure. The angle of incidence is $45^{\circ}$ and the angle of emergence is also $45^{\circ}$. The time spent by the particle in the magnetic field is $...... \, ns$.

$A$ uniform magnetic field $B$ and a uniform electric field $E$ act in a common region. An electron is entering this region of space. The correct arrangement for it to escape undeviated is

$A$ charged particle is released from rest in a region of uniform electric and magnetic fields,which are parallel to each other. The locus of the particle will be

$A$ particle of mass $M$ and positive charge $Q$,moving with a constant velocity $\vec{u}_1 = 4\hat{i} \text{ m/s}$,enters a region of uniform static magnetic field normal to the $x-y$ plane. The region of the magnetic field extends from $x = 0$ to $x = L$ for all values of $y$. After passing through this region,the particle emerges on the other side after $10 \text{ ms}$ with a velocity $\vec{u}_2 = 2(\sqrt{3}\hat{i} + \hat{j}) \text{ m/s}$. The correct statement$(s)$ is (are):
$(A)$ The direction of the magnetic field is $-z$ direction.
$(B)$ The direction of the magnetic field is $+z$ direction.
$(C)$ The magnitude of the magnetic field is $\frac{50\pi M}{3Q}$ units.
$(D)$ The magnitude of the magnetic field is $\frac{100\pi M}{3Q}$ units.

An electron accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

$A$ proton moving with a velocity of $8 \times 10^5 \ m/s$ enters a uniform magnetic field normal to the direction of the magnetic field. If the radius of the circular path of the proton in the magnetic field is $8.3 \ cm$, then the magnitude of the magnetic field is (Charge of proton $= 1.6 \times 10^{-19} \ C$ and mass of the proton $= 1.66 \times 10^{-27} \ kg$) (in $mT$)