A charged particle carrying charge $1\,\mu C$ is moving with velocity $(2 \hat{ i }+3 \hat{ j }+4 \hat{ k })\, ms ^{-1} .$ If an external magnetic field of $(5 \hat{ i }+3 \hat{ j }-6 \hat{ k }) \times 10^{-3}\, T$ exists in the region where the particle is moving then the force on the particle is $\overline{ F } \times 10^{-9} N$. The vector $\overrightarrow{ F }$ is :
$-0.30 \hat{ i }+0.32 \hat{ j }-0.09 \hat{ k }$
$-300 \hat{ i }+320 \hat{ j }-90 \hat{ k }$
$-30 \hat{ i }+32 \hat{ j }-9 \hat{ k }$
$-3.0 \hat{ i }+3.2 \hat{ j }-0.9 \hat{ k }$
An electron accelerated through a potential difference $V$ enters 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 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
Two particles $A$ and $B$ of masses ${m_A}$ and ${m_B}$ respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are ${v_A}$ and ${v_B}$ respectively, and the trajectories are as shown in the figure. Then
As shown in the figure, the uniform magnetic field between the two identical plates is $B$. There is a hole in plate. If through this hole a particle of charge $q$, mass $m$ and energy $E$ enters this magnetic field, then the particle will not collide with the upper plate provided
The region between $y = 0$ and $y = d$ contains a magnetic field $\vec B = B\hat z$ A particle of mass $m$ and charge $q$ enters the region with a velocity $\vec v = v\hat i$. If $d = \frac{{mv}}{{2qB}}$ , the acceleration of the charged particle at the point of its emergence at the other side is