Question

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

• A. its speed will change
• B. its path will change
• C. its speed as well as path will change
• D. it will not experience any force

Hint:

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.

Answer 1

The Correct Option is B

Concept: Magnetic force on a charged particle is \[ \vec F=q(\vec v\times\vec B). \] A magnetic field changes the direction of velocity but not its magnitude.

Step 1: Calculate \(\vec v\times\vec B\). \[ \vec v = 3\hat i+4\hat j \] \[ \vec B = 5\hat k \] \[ \vec v\times\vec B = \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & 4 & 0\\ 0 & 0 & 5 \end{vmatrix} \] \[ = 20\hat i-15\hat j. \] Hence \[ \vec F=q(20\hat i-15\hat j)\neq0. \]

Step 2: Interpret the result. Since force is non-zero, \[ \text{path changes}. \] Also, \[ \vec F\perp\vec v, \] so the magnetic force does no work. Therefore, \[ \text{speed remains constant}. \] \[\begin{aligned} \boxed{\text{Its path will change}} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.