A charged particle moves in a magnetic field `vecB=10hati` with initial velocity `vecu=5veci+4hatj`. The path of the particle will be

Options

A. Straight line
B. circle
C. helical
D. None of these

Explanation

To determine the path of a charged particle moving in a magnetic field, we can follow these steps:

### Step 1: Identify the Given Values
- Magnetic field \( \vec{B} = 10 \hat{i} \) (in the x-direction)
- Initial velocity \( \vec{u} = 5 \hat{i} + 4 \hat{j} \)

### Step 2: Calculate the Magnetic Force
The magnetic force \( \vec{F} \) acting on a charged particle is given by the equation:
\[ \vec{F} = q (\vec{u} \times \vec{B}) \]
where \( q \) is the charge of the particle.

### Step 3: Compute the Cross Product \( \vec{u} \times \vec{B} \)
To compute the cross product, we can set up the determinant:
\[ \vec{u} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 4 & 0 \\ 10 & 0 & 0 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 4 & 0 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 5 & 0 \\ 10 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 5 & 4 \\ 10 & 0 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 4 & 0 \\ 0 & 0 \end{vmatrix} = 0 \)
2. \( \begin{vmatrix} 5 & 0 \\ 10 & 0 \end{vmatrix} = 0 \)
3. \( \begin{vmatrix} 5 & 4 \\ 10 & 0 \end{vmatrix} = (5)(0) - (4)(10) = -40 \)

Putting it all together:
\[ \vec{u} \times \vec{B} = 0 \hat{i} - 0 \hat{j} - 40 \hat{k} = -40 \hat{k} \]

### Step 4: Determine the Direction of Motion
The magnetic force \( \vec{F} \) is directed along the \( -\hat{k} \) direction (out of the plane). This indicates that the particle will experience a force perpendicular to both its velocity and the magnetic field.

### Step 5: Analyze the Path of the Particle
Since the magnetic force is always perpendicular to the velocity of the charged particle, the particle will not move in a straight line. Instead, it will undergo circular motion in the plane perpendicular to the magnetic field.

### Conclusion
The path of the charged particle will be **circular** in the plane perpendicular to the magnetic field \( \vec{B} \).

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