Question

A charge \( q \) moving with velocity \( v \) in a magnetic field of induction \( B \), experiences force \( F \). The angle between \( v \) and \( B \) is \( \theta \). The speed of \( q \) after one second will be

• A. \( \frac{v}{B} \)
• B. \( v \)
• C. \( v \times B \)
• D. \( \frac{B}{v} \)

Hint:

In a magnetic field, the force on a moving charge only changes the direction of the velocity, not its magnitude. Therefore, the speed remains constant.

Answer 1

The Correct Option is B

Step 1: Understanding the force on a moving charge.
The force on a moving charge in a magnetic field is given by: \[ F = qvB \sin(\theta) \] where \( F \) is the force on the charge, \( q \) is the charge, \( v \) is the velocity of the charge, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field.
Step 2: Force and velocity relationship.
The force \( F \) causes the charge to move in a circular or spiral path, but it does not change the speed of the charge. Therefore, after one second, the speed of the charge will remain constant as \( v \).
Step 3: Conclusion.
Thus, the speed of the charge after one second will be \( v \), corresponding to option (B).