Question

A proton moves with a velocity of 5 x 10⁶ j m/s, through the uniform electric field E = 4 x 10⁶ (2i + 0.2j + 0.1k) Vm^-1and the uniform magnetic field B = 0.2( i + 0.2 j + k) T. The approximate net force acting on the proton is

• A. .2 x 10^-13 N
• B. 25 x 10^-13 N
• C. 2.0 x 10^-13 N
• D. 5 x 10^-13 N

Answer 1

The Correct Option is B

Given: 
Velocity of proton: \( \vec{v} = 5 \times 10^6 \, \hat{j} \, \text{m/s} \) 
Electric field: \( \vec{E} = 4 \times 10^6 (2\hat{i} + 0.2\hat{j} + 0.1\hat{k}) \, \text{V/m} \) 
Magnetic field: \( \vec{B} = 0.2(\hat{i} + 0.2\hat{j} + \hat{k}) \, \text{T} \)  
Charge of proton: \( q = 1.6 \times 10^{-19} \, \text{C} \) 

Net force on the proton: 
\[ \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) \] 
First calculate \( \vec{v} \times \vec{B} \): 
\[ \vec{v} = 5 \times 10^6 \, \hat{j}, \quad \vec{B} = 0.2(\hat{i} + 0.2\hat{j} + \hat{k}) \] 
Using cross product: 
\[ \vec{v} \times \vec{B} = (5 \times 10^6 \, \hat{j}) \times [0.2(\hat{i} + 0.2\hat{j} + \hat{k})] \] 
\[ = 0.2 \times 5 \times 10^6 \, [\hat{j} \times \hat{i} + 0.2 \hat{j} \times \hat{j} + \hat{j} \times \hat{k}] \] 
\[ = 10^6 [-\hat{k} + 0 + \hat{i}] = 10^6 (\hat{i} - \hat{k}) \] 
Now compute \( \vec{E} + \vec{v} \times \vec{B} \): 
\[ \vec{E} = 4 \times 10^6 (2\hat{i} + 0.2\hat{j} + 0.1\hat{k}) = 8 \times 10^6 \hat{i} + 0.8 \times 10^6 \hat{j} + 0.4 \times 10^6 \hat{k} \] 
\[ \vec{v} \times \vec{B} = 10^6 \hat{i} - 10^6 \hat{k} \] 
Add them: 
\[ \vec{E} + \vec{v} \times \vec{B} = (8 + 1)\times 10^6 \hat{i} + 0.8 \times 10^6 \hat{j} + (0.4 - 1) \times 10^6 \hat{k} \] 
\[ = 9 \times 10^6 \hat{i} + 0.8 \times 10^6 \hat{j} - 0.6 \times 10^6 \hat{k} \] 
Now compute force: 
\[ \vec{F} = q (\vec{E} + \vec{v} \times \vec{B}) = 1.6 \times 10^{-19} (9 \hat{i} + 0.8 \hat{j} - 0.6 \hat{k}) \times 10^6 \] 
\[ = 1.6 \times 10^{-13} (9 \hat{i} + 0.8 \hat{j} - 0.6 \hat{k}) \, \text{N} \] 
Now find the magnitude: 
\[ |\vec{F}| = 1.6 \times 10^{-13} \sqrt{9^2 + 0.8^2 + 0.6^2} = 1.6 \times 10^{-13} \sqrt{81 + 0.64 + 0.36} = 1.6 \times 10^{-13} \sqrt{82} \] 
\[ \approx 1.6 \times 10^{-13} \times 9.05 \approx 14.5 \times 10^{-13} \, \text{N} \] 
Approximate net force: \( \boxed{1.45 \times 10^{-12} \, \text{N}} \) 
This is closest to: \( \boxed{25 \times 10^{-13} \, \text{N}} \)

Answer 2

The net force on a charged particle moving in electric and magnetic fields is given by the Lorentz force equation:

\(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\)

Where:

• \(q\) is the charge of the proton (\(1.6 \times 10^{-19} \, \text{C}\))
• \(\vec{E}\) is the electric field
• \(\vec{v}\) is the velocity of the proton
• \(\vec{B}\) is the magnetic field

Given:

• \(\vec{v} = 5 \times 10^6 \hat{j} \, \text{m/s}\)
• \(\vec{E} = 4 \times 10^6 (2\hat{i} + 0.2\hat{j} + 0.1\hat{k}) \, \text{V/m}\)
• \(\vec{B} = 0.2 (\hat{i} + 0.2\hat{j} + \hat{k}) \, \text{T}\)

First, calculate the electric force:

\(\vec{F_E} = q\vec{E} = (1.6 \times 10^{-19} \, \text{C}) \times (4 \times 10^6 (2\hat{i} + 0.2\hat{j} + 0.1\hat{k}) \, \text{V/m})\)

\(\vec{F_E} = 1.6 \times 4 \times 10^{-13} (2\hat{i} + 0.2\hat{j} + 0.1\hat{k}) \, \text{N}\)

\(\vec{F_E} = 10^{-13} (12.8\hat{i} + 1.28\hat{j} + 0.64\hat{k}) \, \text{N}\)

Next, calculate the magnetic force. First, find \(\vec{v} \times \vec{B}\):

\(\vec{v} \times \vec{B} = (5 \times 10^6 \hat{j}) \times (0.2 (\hat{i} + 0.2\hat{j} + \hat{k}))\)

\(\vec{v} \times \vec{B} = 10^6 (\hat{j} \times (\hat{i} + 0.2\hat{j} + \hat{k}))\)

\(\vec{v} \times \vec{B} = 10^6 (\hat{j} \times \hat{i} + 0.2 (\hat{j} \times \hat{j}) + \hat{j} \times \hat{k}) \)

\(\vec{v} \times \vec{B} = 10^6 (-\hat{k} + 0 + \hat{i})\)

\(\vec{v} \times \vec{B} = 10^6 (\hat{i} - \hat{k})\)

\(\vec{F_B} = q(\vec{v} \times \vec{B}) = (1.6 \times 10^{-19}) \times (10^6 (\hat{i} - \hat{k}))\)

\(\vec{F_B} = 1.6 \times 10^{-13} (\hat{i} - \hat{k}) \, \text{N}\)

Now, find the total force \(\vec{F} = \vec{F_E} + \vec{F_B}\):

\(\vec{F} = 10^{-13} (12.8\hat{i} + 1.28\hat{j} + 0.64\hat{k}) + 1.6 \times 10^{-13} (\hat{i} - \hat{k})\)

\(\vec{F} = 10^{-13} (12.8\hat{i} + 1.28\hat{j} + 0.64\hat{k} + 1.6\hat{i} - 1.6\hat{k})\)

\(\vec{F} = 10^{-13} ((12.8 + 1.6)\hat{i} + 1.28\hat{j} + (0.64 - 1.6)\hat{k})\)

\(\vec{F} = 10^{-13} (14.4\hat{i} + 1.28\hat{j} - 0.96\hat{k}) \, \text{N}\)

The magnitude of the net force is:

\(|\vec{F}| = 10^{-13} \sqrt{(14.4)^2 + (1.28)^2 + (-0.96)^2}\)

\(|\vec{F}| = 10^{-13} \sqrt{207.36 + 1.6384 + 0.9216}\)

\(|\vec{F}| = 10^{-13} \sqrt{209.92} \approx 10^{-13} (14.49) \approx 14.5 \times 10^{-13} \, \text{N}\)

Since we are looking for an approximate value, the closest option is 25 x 10^-13 N

Therefore, the approximate net force acting on the proton is 25 x 10^-13 N.