Question

Write Fleming's left hand rule. A particle of having charge \( 1.0 \times 10^{-9} \, \text{C} \) enters in a magnetic field of \( \vec{B} = (4 \hat{i} + 3 \hat{j}) \, \text{tesla} \) with velocity \( \vec{v} = (-75 \hat{i} + 100 \hat{j}) \, \text{m/s} \). Find the magnitude and direction of magnetic force exerting on the particle.

Hint:

Fleming's Left Hand Rule helps you determine the direction of the force on a charged particle moving in a magnetic field.

Answer 1

Step 1: Fleming's Left Hand Rule.
Fleming's Left Hand Rule is used to find the direction of force exerted on a moving charged particle in a magnetic field. It states that if you stretch the thumb, index finger, and middle finger of your left hand at right angles to each other: - The index finger points in the direction of the magnetic field \( \vec{B} \), - The middle finger points in the direction of the velocity \( \vec{v} \) of the particle, - The thumb will point in the direction of the force \( \vec{F} \) on the charged particle.
Step 2: Formula for magnetic force.
The force \( \vec{F} \) on a charged particle moving in a magnetic field is given by the Lorentz force law: \[ \vec{F} = q (\vec{v} \times \vec{B}) \] where: - \( q \) is the charge of the particle, - \( \vec{v} \) is the velocity of the particle, - \( \vec{B} \) is the magnetic field, - \( \times \) denotes the vector cross product.
Step 3: Cross product calculation.
We are given: \[ \vec{v} = (-75 \hat{i} + 100 \hat{j}) \, \text{m/s}, \quad \vec{B} = (4 \hat{i} + 3 \hat{j}) \, \text{tesla}, \quad q = 1.0 \times 10^{-9} \, \text{C} \] To find the magnetic force, we calculate the cross product \( \vec{v} \times \vec{B} \): \[ \vec{v} \times \vec{B} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ -75 & 100 & 0 \ 4 &\ 3 & 0 \end{matrix} \right| \] Expanding this determinant: \[ \vec{v} \times \vec{B} = \hat{i}(100 \times 0 - 0 \times 3) - \hat{j}(-75 \times 0 - 0 \times 4) + \hat{k}(-75 \times 3 - 100 \times 4) \] \[ \vec{v} \times \vec{B} = \hat{k}(-225 - 400) = \hat{k}(-625) \] Thus: \[ \vec{v} \times \vec{B} = -625 \hat{k} \, \text{N C}^{-1} \]
Step 4: Force on the particle.
Now, we use the formula for the force: \[ \vec{F} = q (\vec{v} \times \vec{B}) = 1.0 \times 10^{-9} \times (-625 \hat{k}) \] \[ \vec{F} = -625 \times 10^{-9} \hat{k} = -0.625 \times 10^{-6} \hat{k} \, \text{N} \]
Step 5: Conclusion.
The magnitude of the force is \( 0.625 \, \mu\text{N} \), and the direction is along the negative \( \hat{k} \)-axis (perpendicular to both \( \vec{v} \) and \( \vec{B} \)).