A non-relativistic proton beam passes without deviation through a region of space where there are uniform transverse mutually perpendicular electric and magnetic fields with `E= 120 kVm^-1 and B=50 mT.` Then the beam strikes a grounded target. Find the force which the beam acts on the target if the beam current is equal to `i=0.8 mA`.
Mass of protons `= 1.67xx10^(-27) kg`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the conditions for the proton beam
The proton beam passes through a region with electric field \( E \) and magnetic field \( B \) without deviation. This means that the force due to the electric field is equal to the force due to the magnetic field.

### Step 2: Set up the equation for force balance
The force due to the electric field \( F_E \) is given by:
\[ F_E = qE \]
The force due to the magnetic field \( F_B \) is given by:
\[ F_B = qvB \]
Setting these equal gives:
\[ qE = qvB \]
We can cancel \( q \) (the charge of the proton) from both sides (as long as \( q \neq 0 \)):
\[ E = vB \]

### Step 3: Solve for the velocity \( v \)
Rearranging the equation gives:
\[ v = \frac{E}{B} \]
Substituting the given values \( E = 120 \times 10^3 \, \text{V/m} \) and \( B = 50 \times 10^{-3} \, \text{T} \):
\[ v = \frac{120 \times 10^3}{50 \times 10^{-3}} = 2.4 \times 10^6 \, \text{m/s} \]

### Step 4: Calculate the number of protons hitting the target per second
The current \( I \) is related to the charge \( q \) and the number of protons \( n \) hitting the target per second:
\[ I = nq \]
Where \( q \) (the charge of a proton) is approximately \( 1.6 \times 10^{-19} \, \text{C} \). Rearranging gives:
\[ n = \frac{I}{q} \]
Substituting \( I = 0.8 \times 10^{-3} \, \text{A} \):
\[ n = \frac{0.8 \times 10^{-3}}{1.6 \times 10^{-19}} = 5 \times 10^{15} \, \text{protons/s} \]

### Step 5: Calculate the mass of protons hitting the target per second
The mass of the protons hitting the target per second is given by:
\[ \text{mass} = n \cdot m_p \]
Where \( m_p \) is the mass of a proton \( = 1.67 \times 10^{-27} \, \text{kg} \):
\[ \text{mass} = 5 \times 10^{15} \cdot 1.67 \times 10^{-27} = 8.35 \times 10^{-12} \, \text{kg} \]

### Step 6: Calculate the force acting on the target
The force \( F \) can be calculated using the change in momentum per unit time:
\[ F = \frac{\Delta p}{\Delta t} = \frac{m \cdot v}{t} \]
Since we are considering 1 second:
\[ F = \text{mass} \cdot v \]
Substituting the values:
\[ F = 8.35 \times 10^{-12} \cdot 2.4 \times 10^6 = 20 \times 10^{-6} \, \text{N} = 20 \, \mu\text{N} \]

### Final Answer
The force with which the beam acts on the target is \( 20 \, \mu\text{N} \).

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