The figure shows a region of length $'l'$ with a uniform magnetic field of $0.3\, T$ in it and a proton entering the region with velocity $4 \times 10^{5}\, ms ^{-1}$ making an angle $60^{\circ}$ with the field. If the proton completes $10$ revolution by the time it cross the region shown, $l$ is close to....... $m$

(mass of proton $=1.67 \times 10^{-27} \,kg ,$ charge of the proton $\left.=1.6 \times 10^{-19}\, C \right)$

An electron enters the space between the plates of a charged capacitor as shown. The charge density on the plate is $\sigma $. Electric intensity in the space between the plates is $E$. A uniform magnetic field $B$ also exists in that space perpendicular to the direction of $E$. The electron moves perpendicular to both $\vec E$ and $\vec B$ without any change in direction. The time taken by the electron to travel a distance $\ell $ is the space is

An electron is accelerated by a potential difference of $12000\, volts$. It then enters a uniform magnetic field of ${10^{ - 3}}\,T$ applied perpendicular to the path of electron. Find the radius of path. Given mass of electron $ = 9 \times {10^{ - 31}}\,kg$ and charge on electron $ = 1.6 \times {10^{ - 19}}\,C$

A charged particle is moving with velocity $v$ in a magnetic field of induction $B$. The force on the particle will be maximum when

An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then :

An electric field of $1500\, V/m$ and a magnetic field of $0.40\, weber/metre^2$ act on  a moving electron. The minimum uniform speed along a straight line the electron could  have is