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

Answer 1

The Correct Option is B

A proton is moving through both an electric field (\( \vec{E} \)) and a magnetic field (\( \vec{B} \)). The total force on the proton is given by the Lorentz force law:

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

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

We are 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, let's calculate the cross product \( \vec{v} \times \vec{B} \):

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

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

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

Now, let's calculate the total electric field force:

\( q\vec{E} = 1.6 \times 10^{-19} \times 4 \times 10^6 (2\hat{i} + 0.2\hat{j} + 0.1\hat{k}) = 6.4 \times 10^{-13} (2\hat{i} + 0.2\hat{j} + 0.1\hat{k})\)

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

Next, let's calculate the magnetic force:

\( q(\vec{v} \times \vec{B}) = 1.6 \times 10^{-19} \times 10^6 (\hat{i} - \hat{k}) = 1.6 \times 10^{-13} (\hat{i} - \hat{k}) = (1.6\hat{i} - 1.6\hat{k}) \times 10^{-13} \, \text{N}\)

Finally, let's add the electric and magnetic forces:

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

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

To approximate the net force's magnitude, we find the magnitude of the components:

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

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

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

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

Therefore, the approximate net force acting on the proton is: \( \approx 5 \times 10^{-13} \, \text{N} \).