Question

A charged particle moving with a velocity \(\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}\) in a magnetic field \(\vec{B}\) experiences a force \(\vec{F} = F_1\hat{i} + F_2\hat{j}\). Here \(v_1, v_2, F_1, F_2\) are all constants. Then \(\vec{B}\) can be:

• A. \(\vec{B} = B_1\hat{i} + B_2\hat{j}\) with \(\frac{v_1}{v_2} = \frac{B_1}{B_2}\),
• B. \(\vec{B} = B_1\hat{i} + B_2\hat{j} + B_3\hat{k}\) with \(\frac{v_1}{v_2} = \frac{B_1}{B_2}\),
• C. \(\vec{B} = B_3\hat{j}\) with \(B_1 = B_2 = 0\),
• D. \(\vec{B} = B_1\hat{j} + B_2\hat{k}\) with \(\vec{B} = B_1\hat{i} + B_2\hat{j} + B_3\hat{k}\) and \(\frac{B_1}{B_2} = \frac{v_1}{v_2}\).

Hint:

Use the Lorentz force law to relate the velocity and magnetic field components, ensuring the force components match.

Answer 1

The Correct Option is B

Step 1: Lorentz Force
The force on a moving charged particle with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is given by the Lorentz force law:
\[\vec{F} = q \vec{v} \times \vec{B},\]
where \(q\) is the particle's charge. Because the force is \(\vec{F} = F_1\hat{i} + F_2\hat{j}\), the force's components are derived from the cross product of \(\vec{v}\) and \(\vec{B}\).
Step 2: Calculate the Cross Product \(\vec{v} \times \vec{B}\)
\[\vec{v} \times \vec{B} =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\v_1 & v_2 & v_3 \\B_1 & B_2 & B_3\end{vmatrix}.\]
Expanding the determinant yields:
\[\vec{v} \times \vec{B} = (v_2B_3 - v_3B_2)\hat{i} - (v_1B_3 - v_3B_1)\hat{j} + (v_1B_2 - v_2B_1)\hat{k}.\]
Step 3: Component Comparison
Comparing the force components \(\vec{F} = F_1\hat{i} + F_2\hat{j}\) with the cross product result, we find:
\[F_1 = v_2B_3 - v_3B_2, \quad F_2 = -(v_1B_3 - v_3B_1).\]
These equations imply a relationship between \(\vec{v}\) and \(\vec{B}\)'s components, specifically:
\[\frac{v_1}{v_2} = \frac{B_1}{B_2}.\]
Step 4: Conclusion
Therefore, the magnetic field \(\vec{B}\) must have components in all three directions \(\hat{i}, \hat{j}, \hat{k}\), and \(\vec{B}\) is expressed as:
\[\vec{B} = B_1\hat{i} + B_2\hat{j} + B_3\hat{k}, \quad \text{with } \frac{v_1}{v_2} = \frac{B_1}{B_2}.\]