Question

A proton is projected with velocity $\overrightarrow{v}=2\hat{i}$ m/s in a region where magnetic field is $\hat{B}=(\hat{i}+3\hat{j}+4\hat{k})\mu T$ and electric field, $\hat{E}=10\hat{i} \mu Vm^{-1}$. Then, find out the net approximate acceleration of proton. (mass of proton = $1.6 \times 10^{-27}$ kg)

Answer 1

The net approximate acceleration of the proton is $1000\sqrt{2}$ m/s$^2$ (or approximately 1414 m/s$^2$).

Answer 2

The problem asks for the net approximate acceleration of a proton moving in combined electric and magnetic fields.

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

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

According to Newton's second law, the acceleration of the particle is:

$\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{q}{m}(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})$

Given values:

• Charge of proton, $q = +e = 1.6 \times 10^{-19}$ C
• Mass of proton, $m = 1.6 \times 10^{-27}$ kg
• Velocity, $\overrightarrow{v} = 2\hat{i}$ m/s
• Magnetic field, $\overrightarrow{B} = (\hat{i}+3\hat{j}+4\hat{k})\mu T = (\hat{i}+3\hat{j}+4\hat{k}) \times 10^{-6}$ T
• Electric field, $\overrightarrow{E} = 10\hat{i} \mu Vm^{-1} = 10\hat{i} \times 10^{-6}$ Vm$^{-1}$

1. Calculate the cross product $\overrightarrow{v} \times \overrightarrow{B}$:

$\overrightarrow{v} \times \overrightarrow{B} = (2\hat{i}) \times ((\hat{i}+3\hat{j}+4\hat{k}) \times 10^{-6})$ $= 10^{-6} [(2\hat{i} \times \hat{i}) + (2\hat{i} \times 3\hat{j}) + (2\hat{i} \times 4\hat{k})]$

Using $\hat{i} \times \hat{i} = 0$, $\hat{i} \times \hat{j} = \hat{k}$, and $\hat{i} \times \hat{k} = -\hat{j}$:

$\overrightarrow{v} \times \overrightarrow{B} = 10^{-6} [0 + 6(\hat{i} \times \hat{j}) + 8(\hat{i} \times \hat{k})]$ $= 10^{-6} [6\hat{k} - 8\hat{j}]$ $= 10^{-6} (-8\hat{j} + 6\hat{k})$ T m/s

2. Calculate the term $(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})$:

$\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B} = (10\hat{i} \times 10^{-6}) + (10^{-6} (-8\hat{j} + 6\hat{k}))$ $= 10^{-6} (10\hat{i} - 8\hat{j} + 6\hat{k})$

3. Calculate the acceleration $\overrightarrow{a}$:

$\overrightarrow{a} = \frac{q}{m}(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})$ $\overrightarrow{a} = \frac{1.6 \times 10^{-19}}{1.6 \times 10^{-27}} \times 10^{-6} (10\hat{i} - 8\hat{j} + 6\hat{k})$ $\overrightarrow{a} = 10^{(-19 - (-27))} \times 10^{-6} (10\hat{i} - 8\hat{j} + 6\hat{k})$ $\overrightarrow{a} = 10^8 \times 10^{-6} (10\hat{i} - 8\hat{j} + 6\hat{k})$ $\overrightarrow{a} = 10^2 (10\hat{i} - 8\hat{j} + 6\hat{k})$ $\overrightarrow{a} = 1000\hat{i} - 800\hat{j} + 600\hat{k}$ m/s$^2$

4. Calculate the magnitude of the acceleration:

The "net approximate acceleration" typically refers to the magnitude of the acceleration vector.

$|\overrightarrow{a}| = \sqrt{(1000)^2 + (-800)^2 + (600)^2}$ $|\overrightarrow{a}| = \sqrt{1000000 + 640000 + 360000}$ $|\overrightarrow{a}| = \sqrt{2000000}$ $|\overrightarrow{a}| = \sqrt{2 \times 10^6}$ $|\overrightarrow{a}| = 10^3 \sqrt{2}$ $|\overrightarrow{a}| \approx 1000 \times 1.414$ $|\overrightarrow{a}| \approx 1414$ m/s$^2$

Explanation of the solution:

1. Calculate the Lorentz force acting on the proton, which is the sum of the electric force ($q\overrightarrow{E}$) and the magnetic force ($q(\overrightarrow{v} \times \overrightarrow{B})$).
2. Substitute the given values for charge $q$, velocity $\overrightarrow{v}$, electric field $\overrightarrow{E}$, and magnetic field $\overrightarrow{B}$ into the Lorentz force equation.
3. Perform the vector cross product $\overrightarrow{v} \times \overrightarrow{B}$.
4. Sum the electric field term and the result of the cross product.
5. Divide the net force by the mass of the proton $m$ to find the acceleration $\overrightarrow{a} = \frac{q}{m}(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})$.
6. Calculate the magnitude of the acceleration vector $|\overrightarrow{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}$.