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Step-by-Step Solution
Step 1: Understand the effect of magnetic force on the particle
When a charged particle with charge $Q$, mass $m$, and kinetic energy $T$ enters a uniform magnetic field $ \overrightarrow{B}$ perpendicular to its velocity, it experiences a magnetic force given by $ \overrightarrow{F} = Q( \overrightarrow{v} \times \overrightarrow{B} )$. This force is always perpendicular to the velocity of the particle.
Step 2: Analyze work done by the magnetic force
The magnetic force does no work on the particle because it is perpendicular to the particle’s displacement at every instant of motion. Mathematically, work done $W$ by a force $ \overrightarrow{F}$ is $W = \overrightarrow{F} \cdot \overrightarrow{d}$, which becomes zero if $ \overrightarrow{F}$ is always perpendicular to $ \overrightarrow{d}$.
Step 3: Relate work done to kinetic energy
Since no work is done on the particle by the magnetic field, there is no change in the kinetic energy of the particle. According to the work-energy theorem, the change in kinetic energy is equal to the net work done on the particle. Here, net work done is zero, so the kinetic energy remains constant.
Step 4: Conclude the final kinetic energy
No matter how long the particle remains in the magnetic field, because the force is always perpendicular to its velocity, its kinetic energy remains the same. Therefore, after 3 seconds (or any time interval), the kinetic energy will still be $T$.
Final Answer
The kinetic energy of the particle after 3 seconds is $T$.
