Question

A charged particle having a mass of 100g and charge of 0.1C is projected from the origin with a speed of $\sqrt{10}m/s$ at an angle of 45° with the vertical. When it is at the highest point of its trajectory, the magnitude of magnetic field at the origin will be:

• A. 8.1 x 10-8 T
• B. 1.9 × 10-8 T
• C. 3.2 x 10-8 T
• D. 6.4 × 10-8 T

Answer 1

Answer: (C)

3.2 x 10-8 T

Answer 2

The problem involves calculating the magnetic field at the origin due to a charged particle undergoing projectile motion, specifically when it is at the highest point of its trajectory.

1. Analyze the Projectile Motion:

• Mass of the particle, m = 100 g = 0.1 kg

• Charge of the particle, q = 0.1 C

• Initial speed, u = $\sqrt{10}$ m/s

• Angle with the vertical = 45°. Therefore, the angle with the horizontal (θ) is 90° - 45° = 45°.

• We'll use acceleration due to gravity, g = 10 m/s².

• Initial velocity components:

  • Horizontal component, u_x = u cos(θ) = $\sqrt{10}$ cos(45°) = $\sqrt{10} \times \frac{1}{\sqrt{2}} = \sqrt{5}$ m/s

  • Vertical component, u_y = u sin(θ) = $\sqrt{10}$ sin(45°) = $\sqrt{10} \times \frac{1}{\sqrt{2}} = \sqrt{5}$ m/s

• At the highest point of the trajectory:

  • The vertical component of velocity (v_y) becomes zero.
  • The horizontal component of velocity (v_x) remains constant: v_x = u_x = $\sqrt{5}$ m/s.
  • So, the velocity vector at the highest point is $\vec{v} = \sqrt{5} \hat{i}$ (assuming x is horizontal and y is vertical).

• Time to reach the highest point (t):

  • Using v_y = u_y - gt:
  • 0 = $\sqrt{5}$ - 10t
  • t = $\frac{\sqrt{5}}{10}$ s

• Coordinates of the highest point (x, H):

  • Horizontal position, x = u_x t = $\sqrt{5} \times \frac{\sqrt{5}}{10} = \frac{5}{10} = 0.5$ m
  • Maximum height, H = u_y t - $\frac{1}{2}$gt² = $\sqrt{5} \times \frac{\sqrt{5}}{10} - \frac{1}{2} \times 10 \times (\frac{\sqrt{5}}{10})^2$ = 0.5 - $5 \times \frac{5}{100}$ = 0.5 - 0.25 = 0.25 m
  • The position vector of the particle at the highest point from the origin is $\vec{r} = 0.5 \hat{i} + 0.25 \hat{j}$.

2. Calculate the Magnetic Field using Biot-Savart Law for a Moving Charge:

The magnetic field $\vec{B}$ at the origin due to a point charge q moving with velocity $\vec{v}$ at a position $\vec{r}$ (from the origin to the charge) is given by:

$\vec{B} = \frac{\mu_0}{4\pi} \frac{q (\vec{v} \times \vec{r})}{r^3}$

Where $\frac{\mu_0}{4\pi} = 10^{-7}$ T m/A.

• Calculate the cross product $\vec{v} \times \vec{r}$:

  • $\vec{v} \times \vec{r} = (\sqrt{5} \hat{i}) \times (0.5 \hat{i} + 0.25 \hat{j})$
  • $= (\sqrt{5} \times 0.5)(\hat{i} \times \hat{i}) + (\sqrt{5} \times 0.25)(\hat{i} \times \hat{j})$
  • $= 0 + 0.25\sqrt{5} \hat{k}$

• Calculate the magnitude of the position vector r:

  • $r = |\vec{r}| = \sqrt{(0.5)^2 + (0.25)^2}$
  • $r = \sqrt{0.25 + 0.0625} = \sqrt{0.3125}$
  • $r = \sqrt{\frac{3125}{10000}} = \sqrt{\frac{5 \times 625}{16 \times 625}} = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}$ m

• Substitute values into the magnetic field formula:

  • $B = \frac{\mu_0}{4\pi} \frac{q |\vec{v} \times \vec{r}|}{r^3}$
  • $B = 10^{-7} \times \frac{0.1 \times (0.25\sqrt{5})}{(\frac{\sqrt{5}}{4})^3}$
  • $B = 10^{-7} \times \frac{0.1 \times 0.25\sqrt{5}}{\frac{5\sqrt{5}}{64}}$
  • $B = 10^{-7} \times \frac{0.1 \times 0.25 \times 64}{5}$
  • $B = 10^{-7} \times \frac{0.025 \times 64}{5}$
  • $B = 10^{-7} \times 0.005 \times 64$
  • $B = 10^{-7} \times 0.32$
  • $B = 3.2 \times 10^{-8}$ T

The magnitude of the magnetic field at the origin is $3.2 \times 10^{-8}$ T.