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 Law
The force on a charged particle moving 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 charge of the particle. Since the force is given as \(\vec{F} = F_1\hat{i} + F_2\hat{j}\), the components of the force must come from the cross product of \(\vec{v}\) and \(\vec{B}\).
Step 2: Compute 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, we get:
\[\vec{v} \times \vec{B} = (v_2B_3 - v_3B_2)\hat{i} - (v_1B_3 - v_3B_1)\hat{j} + (v_1B_2 - v_2B_1)\hat{k}.\]
Step 3: Compare Components
Comparing the components of \(\vec{F} = F_1\hat{i} + F_2\hat{j}\) with the expression for \(\vec{v} \times \vec{B}\), we get:
\[F_1 = v_2B_3 - v_3B_2, \quad F_2 = -(v_1B_3 - v_3B_1).\]
These equations imply relationships between the components of \(\vec{v}\) and \(\vec{B}\), specifically:
\[\frac{v_1}{v_2} = \frac{B_1}{B_2}.\]
Step 4: Conclusion
From this, we conclude that the magnetic field \(\vec{B}\) must have components in all three directions \(\hat{i}, \hat{j}, \hat{k}\), and the correct expression for \(\vec{B}\) is:
\[\vec{B} = B_1\hat{i} + B_2\hat{j} + B_3\hat{k}, \quad \text{with } \frac{v_1}{v_2} = \frac{B_1}{B_2}.\]

Answer 2

Using the hint provided, we have two conditions based on the fact that the magnetic force is always perpendicular to both the velocity and the magnetic field:

1. $\vec{F} \cdot \vec{v} = 0$
2. $\vec{F} \cdot \vec{B} = 0$

Given $\vec{F} = F_1\hat{i} + F_2\hat{j}$ and $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$, the first condition gives:

$(F_1\hat{i} + F_2\hat{j}) \cdot (v_1\hat{i} + v_2\hat{j} + v_3\hat{k}) = 0$

$F_1v_1 + F_2v_2 + 0 \cdot v_3 = 0$

$F_1v_1 + F_2v_2 = 0$

If $F_2 \neq 0$ and $v_1 \neq 0$, we can write $\frac{F_1}{F_2} = -\frac{v_2}{v_1}$, which matches the first part of the hint (I).

Now, let $\vec{B} = B_1\hat{i} + B_2\hat{j} + B_3\hat{k}$. The second condition gives:

$\vec{F} \cdot \vec{B} = (F_1\hat{i} + F_2\hat{j}) \cdot (B_1\hat{i} + B_2\hat{j} + B_3\hat{k}) = 0$

$F_1B_1 + F_2B_2 + 0 \cdot B_3 = 0$

$F_1B_1 + F_2B_2 = 0$

If $F_2 \neq 0$ and $B_1 \neq 0$, we can write $\frac{F_1}{F_2} = -\frac{B_2}{B_1}$, which matches the second part of the hint (II).

From (I) and (II), we have:

$-\frac{v_2}{v_1} = -\frac{B_2}{B_1}$

$\frac{B_1}{B_2} = \frac{v_1}{v_2}$

This means that the components of the magnetic field in the xy-plane must be proportional to the corresponding components of the velocity in the xy-plane. So, $B_1 = k v_1$ and $B_2 = k v_2$ for some constant $k$. The $z$-component of the magnetic field, $B_3$, is not constrained by these two dot product conditions.

Therefore, the magnetic field $\vec{B}$ can be written as:

$\vec{B} = k v_1 \hat{i} + k v_2 \hat{j} + B_3 \hat{k}$

$\vec{B} = k (v_1 \hat{i} + v_2 \hat{j}) + B_3 \hat{k}$

Now let's check the provided options from the original problem with this form of $\vec{B}$. Since the options were not provided in the original question, we will consider a general form that satisfies the conditions.

A possible magnetic field is one that has its x and y components proportional to $v_1$ and $v_2$ respectively, and any z component. For example, if we choose $k=1$ and $B_3=0$, then $\vec{B} = v_1 \hat{i} + v_2 \hat{j}$.

Let's verify if $\vec{F}$ is perpendicular to this $\vec{B}$:

$\vec{F} \cdot \vec{B} = (F_1\hat{i} + F_2\hat{j}) \cdot (v_1\hat{i} + v_2\hat{j}) = F_1v_1 + F_2v_2$

From the condition $\vec{F} \cdot \vec{v} = 0$, we know that $F_1v_1 + F_2v_2 = 0$. So, $\vec{F} \cdot \vec{B} = 0$ is satisfied.

Therefore, a possible magnetic field is $\vec{B} = v_1 \hat{i} + v_2 \hat{j}$. In general, $\vec{B} = c(v_1 \hat{i} + v_2 \hat{j}) + d \hat{k}$ where $c$ and $d$ are any constants.

Final Answer: The final answer is $B=B_1i+B_2j+B_3k \text{ where } \frac{B_1}{B_2} = \frac{v_1}{v_2}$