A beam of the charged particles is passed through a magnetic field. The work done on the beam by the field is

Options

A. zero
B. independent on the speed of the beam
C. dependent on the deflection of the beam
D. dependent on the magnetic induction `vecB`

Answer

To solve the problem of finding the work done on a beam of charged particles passing through a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Work Done**:
The work done \( W \) by a force is defined as:
\[ W = \vec{F} \cdot \vec{S} \]
where \( \vec{F} \) is the force acting on the charged particles and \( \vec{S} \) is the displacement of the particles.

2. **Magnetic Force on Charged Particles**:
The magnetic force \( \vec{F} \) on a charged particle moving in a magnetic field is given by the Lorentz force equation:
\[ \vec{F} = q (\vec{v} \times \vec{B}) \]
where \( q \) is the charge of the particle, \( \vec{v} \) is the velocity vector of the particle, and \( \vec{B} \) is the magnetic field vector.

3. **Direction of Force**:
The direction of the magnetic force is perpendicular to both the velocity of the charged particles and the magnetic field. This means that the force does not do any work on the charged particles because work is only done when the force has a component in the direction of displacement.

4. **Calculating Work Done**:
Since the magnetic force is perpendicular to the displacement, the angle \( \theta \) between the force and the displacement is \( 90^\circ \). Therefore, we can express the work done as:
\[ W = F \cdot S \cdot \cos(90^\circ) \]
Since \( \cos(90^\circ) = 0 \), we have:
\[ W = F \cdot S \cdot 0 = 0 \]

5. **Conclusion**:
The work done on the beam of charged particles by the magnetic field is zero.

### Final Answer:
The work done on the beam by the magnetic field is **0**.

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