There is a constant homogeneous electric field of `100 Vm^-1` within the region `x=0 and x=0.167 m` pointing in x-direction. There is a constant homogeneous mangetic field B within the region `x=0.167 m and x=0.334 m` pointing in the z-direction. A proton at rest at the origin is released in positive x-direction. Find the minimum strength of the magnetic field B, so that the proton is detected back at `x=0, y=0.167 m.` (mass of proton `= 1.67 xx 10^(-27) kg)`

Explanation

To solve the problem, we need to analyze the motion of a proton in the given electric and magnetic fields. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
- There is a constant electric field \( E = 100 \, \text{V/m} \) in the region \( x = 0 \) to \( x = 0.167 \, \text{m} \).
- There is a constant magnetic field \( B \) in the region \( x = 0.167 \, \text{m} \) to \( x = 0.334 \, \text{m} \) pointing in the z-direction.
- The proton is released from rest at the origin and moves in the positive x-direction.
- The mass of the proton \( m = 1.67 \times 10^{-27} \, \text{kg} \).
- We need to find the minimum strength of the magnetic field \( B \) such that the proton is detected back at \( (x=0, y=0.167) \).
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