Question:easy

A proton has a velocity \[ \vec v=3\hat i+4\hat j\ \text{m s}^{-1} \] and is subjected to a magnetic field \[ \vec B=5\hat k\ \text{T}. \] Then

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Magnetic force \[ \vec F=q(\vec v\times\vec B) \] is always perpendicular to velocity. Therefore it changes only the direction of motion, not the speed.
Updated On: Jun 16, 2026
  • its speed will change
  • its path will change
  • its speed as well as path will change
  • it will not experience any force
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The Correct Option is B

Solution and Explanation

Step 1: The force on a moving charge.
A charge in a magnetic field feels \[ \vec F=q(\vec v\times\vec B). \] The direction of this force decides what happens to the motion.

Step 2: Note the directions given.
Velocity $\vec v=3\hat i+4\hat j$ lies in the $x$-$y$ plane, while $\vec B=5\hat k$ points along $z$. They are at right angles, so the force is non-zero.

Step 3: A useful general fact.
The magnetic force is always perpendicular to $\vec v$ (since $\vec v\times\vec B$ is perpendicular to $\vec v$). A force perpendicular to velocity does no work.

Step 4: Effect on speed.
No work means no change in kinetic energy, so the speed stays exactly the same.

Step 5: Effect on direction.
Even though speed is constant, the sideways force keeps bending the velocity. So the direction, and hence the path, keeps changing.

Step 6: Combine the two results.
Speed unchanged, but path curved. The correct statement is that only the path changes.
\[ \boxed{\text{Its path will change (speed stays the same)}} \]
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