A charged particle of charge 4 mC enters a uniform magnetic field of induction `vecB = 3bari+6barj+6bark` tesla with a velocity `barv = 4bari-xbarj+y bark`. If the particles continues to move undeviated, then the magnitude of velocity of the particle is

Options

A. 10 m/s
B. 15 m/s
C. 12 m/s
D. 8 m/s

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the Problem
We have a charged particle with a charge of \( q = 4 \, \text{mC} = 4 \times 10^{-3} \, \text{C} \) moving in a magnetic field \( \vec{B} = 3\hat{i} + 6\hat{j} + 6\hat{k} \, \text{T} \) with a velocity \( \vec{v} = 4\hat{i} - x\hat{j} + y\hat{k} \). The particle moves undeviated, which implies that the magnetic force acting on it is zero.

### Step 2: Magnetic Force Equation
The magnetic force \( \vec{F} \) on a charged particle is given by:
\[ \vec{F} = q (\vec{v} \times \vec{B}) \]
Since the particle moves undeviated, we have:
\[ \vec{F} = 0 \implies \vec{v} \times \vec{B} = 0 \]

### Step 3: Calculate the Cross Product
To find the values of \( x \) and \( y \), we need to compute the cross product \( \vec{v} \times \vec{B} \):
\[ \vec{v} = 4\hat{i} - x\hat{j} + y\hat{k} \]
\[ \vec{B} = 3\hat{i} + 6\hat{j} + 6\hat{k} \]
The cross product can be calculated using the determinant:
\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -x & y \\ 3 & 6 & 6 \end{vmatrix} \]

### Step 4: Calculate the Determinant
Calculating the determinant:
\[ \vec{v} \times \vec{B} = \hat{i} \left((-x)(6) - (y)(6)\right) - \hat{j} \left(4(6) - (y)(3)\right) + \hat{k} \left(4(6) - (-x)(3)\right) \]
This simplifies to:
\[ = \hat{i} (-6x - 6y) - \hat{j} (24 - 3y) + \hat{k} (24 + 3x) \]

### Step 5: Set Each Component to Zero
For the cross product to be zero, each component must equal zero:
1. \( -6x - 6y = 0 \) (1)
2. \( 24 - 3y = 0 \) (2)
3. \( 24 + 3x = 0 \) (3)

### Step 6: Solve the Equations
From equation (2):
\[ 24 - 3y = 0 \implies 3y = 24 \implies y = 8 \]
From equation (3):
\[ 24 + 3x = 0 \implies 3x = -24 \implies x = -8 \]

### Step 7: Calculate the Magnitude of Velocity
Now we substitute \( x \) and \( y \) back into the velocity vector:
\[ \vec{v} = 4\hat{i} - (-8)\hat{j} + 8\hat{k} = 4\hat{i} + 8\hat{j} + 8\hat{k} \]
The magnitude of the velocity \( |\vec{v}| \) is given by:
\[ |\vec{v}| = \sqrt{(4)^2 + (-8)^2 + (8)^2} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12 \, \text{m/s} \]

### Final Answer
The magnitude of the velocity of the particle is \( 12 \, \text{m/s} \).

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