The radius of curvature of the path of the charged particle in a uniform magnetic field is directly proportional to
The charge on the particle
The momentum of the particle
The energy of the particle
The intensity of the field
A proton is moving along $Z$-axis in a magnetic field. The magnetic field is along $X$-axis. The proton will experience a force along
A charge particle is moving in a uniform magnetic field $(2 \hat{i}+3 \hat{j}) T$. If it has an acceleration of $(\alpha \hat{i}-4 \hat{j}) m / s ^{2}$, then the value of $\alpha$ will be.
Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields $\vec{E}=E_0 \hat{j}$ and $\vec{B}=B_0 \hat{j}$. At time $t=0$, this charge has velocity $\nabla$ in the $x$-y plane, making an angle $\theta$ with $x$-axis. Which of the following option$(s)$ is(are) correct for time $t>0$ ?
$(A)$ If $\theta=0^{\circ}$, the charge moves in a circular path in the $x-z$ plane.
$(B)$ If $\theta=0^{\circ}$, the charge undergoes helical motion with constant pitch along the $y$-axis.
$(C)$ If $\theta=10^{\circ}$, the charge undergoes helical motion with its pitch increasing with time, along the $y$-axis.
$(D)$ If $\theta=90^{\circ}$, the charge undergoes linear but accelerated motion along the $y$-axis.
A particle with charge $+Q$ and mass m enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ$. The direction of the motion of the $m$ particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{{QB}}$ . The time spent by the particle in the field will be
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 $m / s ^{2}$)