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A particle of mass $m$, charge $Q$ and kinetic energy $T$ enters a transverse uniform magnetic field of induction $\vec{B}$. After 3 seconds the kinetic energy of the particle will be -
A
$T$
B
$4T$
C
$3T$
D
$2T$
Detailed Solution
The magnetic force on a moving charge is $\vec{F} = Q(\vec{v} \times \vec{B})$.
This force is always perpendicular to both the velocity $\vec{v}$ and the magnetic field $\vec{B}$.
Power delivered by the force: $P = \vec{F}\cdot\vec{v} = 0$, because $\vec{F} \perp \vec{v}$.
Equivalently, the work done in a small displacement is $dW = \vec{F}\cdot d\vec{r} = 0$, because $\vec{F} \perp d\vec{r}$.
A force that does no work cannot change the kinetic energy or the speed of the particle; it only changes the direction of motion (the particle moves on a circular path).
Hence after 3 seconds (or after any time) the kinetic energy is still $T$.
This force is always perpendicular to both the velocity $\vec{v}$ and the magnetic field $\vec{B}$.
Power delivered by the force: $P = \vec{F}\cdot\vec{v} = 0$, because $\vec{F} \perp \vec{v}$.
Equivalently, the work done in a small displacement is $dW = \vec{F}\cdot d\vec{r} = 0$, because $\vec{F} \perp d\vec{r}$.
A force that does no work cannot change the kinetic energy or the speed of the particle; it only changes the direction of motion (the particle moves on a circular path).
Hence after 3 seconds (or after any time) the kinetic energy is still $T$.