A charge q is moving with a velocity `v_(1)=1hati` m/s at a point in a magentic field and experiences a force `F=q[-hatj+1hatk]` N. If the charge is moving with a voloctiy `v_(2)=hatj` m/s at the same point then it experiences a force `F_(2)=q(1hati-1hatk)` N. The magnetic induction B at that point is

Options

A. `(hat(i)+hat(j)+hat(k))Wb//m^(2)`
B. `(hat(i)-hat(j)+hat(k))Wb//m^(2)`
C. `(-hat(i)+hat(j)-hat(k))Wb//m^(2)`
D. `(hat(i)+hat(j)-hat(k))Wb//m^(2)`

Answer

To find the magnetic induction \( \mathbf{B} \) at the point where a charge \( q \) is moving in a magnetic field, we can use the formula for the magnetic force on a charged particle:

\[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \]

where \( \mathbf{F} \) is the magnetic force, \( \mathbf{v} \) is the velocity of the charge, and \( \mathbf{B} \) is the magnetic field.

### Step 1: Analyze the first scenario

Given:
- Velocity \( \mathbf{v_1} = 1 \hat{i} \, \text{m/s} \)
- Force \( \mathbf{F_1} = q (-\hat{j} + \hat{k}) \, \text{N} \)

Using the force equation:

\[ \mathbf{F_1} = q (\mathbf{v_1} \times \mathbf{B}) \]

Substituting the values:

\[ q (-\hat{j} + \hat{k}) = q (1 \hat{i} \times \mathbf{B}) \]

Dividing both sides by \( q \) (assuming \( q \neq 0 \)):

\[ -\hat{j} + \hat{k} = (1 \hat{i} \times \mathbf{B}) \]

### Step 2: Express the magnetic field \( \mathbf{B} \)

Let \( \mathbf{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \).

Calculating the cross product:

\[ \hat{i} \times \mathbf{B} = \hat{i} \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) = B_y \hat{k} - B_z \hat{j} \]

Thus, we have:

\[ -\hat{j} + \hat{k} = B_y \hat{k} - B_z \hat{j} \]

Comparing components, we get two equations:

1. \( -B_z = -1 \) (from the \( \hat{j} \) component)
2. \( B_y = 1 \) (from the \( \hat{k} \) component)

From equation 1, we find:

\[ B_z = 1 \]

### Step 3: Analyze the second scenario

Given:
- Velocity \( \mathbf{v_2} = \hat{j} \, \text{m/s} \)
- Force \( \mathbf{F_2} = q (1 \hat{i} - \hat{k}) \, \text{N} \)

Using the force equation again:

\[ \mathbf{F_2} = q (\mathbf{v_2} \times \mathbf{B}) \]

Substituting the values:

\[ q (1 \hat{i} - \hat{k}) = q (\hat{j} \times \mathbf{B}) \]

Dividing both sides by \( q \):

\[ 1 \hat{i} - \hat{k} = (\hat{j} \times \mathbf{B}) \]

### Step 4: Calculate the cross product for \( \mathbf{v_2} \)

Calculating the cross product:

\[ \hat{j} \times \mathbf{B} = \hat{j} \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) = -B_x \hat{k} + B_z \hat{i} \]

Thus, we have:

\[ 1 \hat{i} - \hat{k} = -B_x \hat{k} + B_z \hat{i} \]

Comparing components, we get two more equations:

1. \( B_z = 1 \) (from the \( \hat{i} \) component)
2. \( -B_x = -1 \) (from the \( \hat{k} \) component)

From equation 1, we confirm:

\[ B_z = 1 \]

From equation 2, we find:

\[ B_x = 1 \]

### Step 5: Compile the results

From both scenarios, we have:

- \( B_x = 1 \)
- \( B_y = 1 \)
- \( B_z = 1 \)

Thus, the magnetic induction \( \mathbf{B} \) at that point is:

\[ \mathbf{B} = 1 \hat{i} + 1 \hat{j} + 1 \hat{k} = \hat{i} + \hat{j} + \hat{k} \]

### Final Answer

The magnetic induction \( \mathbf{B} \) at that point is:

\[ \mathbf{B} = \hat{i} + \hat{j} + \hat{k} \]