A magnet of magnetic moment `20_(hatk) Am^(2)` is placed along the `z-`axis in a magnetic field `vec(B)=(0.4hat(j)+0.5hat(k)) T`. The torque acting on the magnet is

Options

A. `8hat(i)N-m`
B. `6hat(i)N-m`
C. `-8hat(i) N-m`
D. `-6hat(i)N-m`

Explanation

To find the torque acting on the magnet, we will use the formula for torque (\(\vec{\tau}\)) acting on a magnetic moment (\(\vec{m}\)) in a magnetic field (\(\vec{B}\)). The formula is given by:

\[ \vec{\tau} = \vec{m} \times \vec{B} \]

### Step 1: Identify the magnetic moment and magnetic field
The magnetic moment is given as:
\[ \vec{m} = 20 \hat{k} \, \text{Am}^2 \]
The magnetic field is given as:
\[ \vec{B} = 0.4 \hat{j} + 0.5 \hat{k} \, \text{T} \]

### Step 2: Write the cross product
Now we need to calculate the cross product \(\vec{m} \times \vec{B}\):
\[ \vec{\tau} = (20 \hat{k}) \times (0.4 \hat{j} + 0.5 \hat{k}) \]

### Step 3: Distribute the cross product
Using the distributive property of the cross product:
\[ \vec{\tau} = 20 \hat{k} \times (0.4 \hat{j}) + 20 \hat{k} \times (0.5 \hat{k}) \]

### Step 4: Calculate each term
1. **First term**:
\[ 20 \hat{k} \times (0.4 \hat{j}) = 20 \times 0.4 (\hat{k} \times \hat{j}) = 8 (\hat{k} \times \hat{j}) \]
We know that \(\hat{k} \times \hat{j} = -\hat{i}\) (using the right-hand rule).
Thus,
\[ 8 (\hat{k} \times \hat{j}) = 8 (-\hat{i}) = -8 \hat{i} \]

2. **Second term**:
\[ 20 \hat{k} \times (0.5 \hat{k}) = 0 \quad \text{(since the cross product of any vector with itself is zero)} \]

### Step 5: Combine the results
Now, combine the results from the two terms:
\[ \vec{\tau} = -8 \hat{i} + 0 = -8 \hat{i} \]

### Final Result
Thus, the torque acting on the magnet is:
\[ \vec{\tau} = -8 \hat{i} \, \text{Nm} \]