If a proton enters perpendicularly a magnetic field with velocity $v$, then time period of revolution is $T$. If proton enters with velocity $2 v$, then time period will be
$T$
$2 T$
$3 T$
$4 T$
A particle having charge of $10\,\mu C$ and $1\,\mu g$ mass moves along circular path of $10\, cm$ radius in the effect of uniform magnetic field of $0.1\, T$. When charge is at point $'P'$, a uniform electric field applied in the region so charge moves tangentially with constant speed. The value of electric field is......$V/m$
The magnetic moments associated with two closely wound circular coils $A$ and $B$ of radius $r_A=10 cm$ and $r_B=20 cm$ respectively are equal if: (Where $N _A, I _{ A }$ and $N _B, I _{ B }$ are number of turn and current of $A$ and $B$ respectively)
A proton (or charged particle) moving with velocity $v$ is acted upon by electric field $E$ and magnetic field $B$. The proton will move undeflected if
An $e^-$ is moving parallel to a long current carrying wire as shown. Force on electron is
Two ions of masses $4 \,{amu}$ and $16\, amu$ have charges $+2 {e}$ and $+3 {e}$ respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then :