A charged particle begins to move from the origin in a region which has a uniform magnetic field in the x-direction and a uniform electric field in the y-direction. Its speed is v when it reaches the point (x, y, z). Then, v will depend

Options

A. only on x
B. only on y
C. on both x and y, but not z
D. on x,y and z

Explanation

To solve the problem step by step, we need to analyze the motion of a charged particle in the presence of both electric and magnetic fields.

### Step 1: Understanding the Fields
The problem states that there is a uniform magnetic field in the x-direction and a uniform electric field in the y-direction. We can denote these fields as:
- Magnetic Field, **B** = B₀ **i** (where **i** is the unit vector in the x-direction)
- Electric Field, **E** = E₀ **j** (where **j** is the unit vector in the y-direction)

### Step 2: Analyzing the Forces
A charged particle (with charge **Q**) moving in these fields will experience forces due to both fields:
- The force due to the electric field is given by **F_E = Q** **E** = Q E₀ **j**.
- The force due to the magnetic field is given by **F_B = Q** (**v** × **B**). However, since the magnetic field is in the x-direction and the particle's motion is in the y-direction, the magnetic force will not do work on the particle. The magnetic force only changes the direction of the velocity, not its magnitude.

### Step 3: Work Done on the Particle
Since the magnetic force does no work, the work done on the particle is solely due to the electric field. The work done by the electric field when the particle moves a distance **s** in the direction of the electric field is:
- **W = F_E · s = Q E₀ s**.

### Step 4: Relating Work Done to Kinetic Energy
According to the work-energy theorem, the work done on the particle is equal to the change in kinetic energy:
- **W = ΔKE = KE_final - KE_initial**.
Assuming the particle starts from rest, the initial kinetic energy is zero, so:
- **W = KE_final = (1/2) m v²**.

### Step 5: Setting Up the Equation
From the above relationships, we can equate the work done to the change in kinetic energy:
- **Q E₀ s = (1/2) m v²**.

### Step 6: Solving for Velocity
From the equation, we can express the speed **v** in terms of the distance **s**:
- **v² = (2 Q E₀ s) / m**.
- **v = √((2 Q E₀ s) / m)**.

### Step 7: Conclusion
The speed **v** of the charged particle when it reaches the point (x, y, z) depends on the distance **s** it has moved in the direction of the electric field (which is the y-direction).

### Final Answer
Thus, the speed **v** will depend on the distance **s** traveled in the direction of the electric field.

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