A magnet of magnetic moment `50 hat(i) A-m^(2)` is placed along the x-axis in a magnetic field `vec(B)=(0.5hat(i)+3.0hat(j))T`. The torque acting on the magnet is

Options

A. `175 hat k `Nm
B. `75hat i`Nm
C. `150hat k`Nm
D. `25sqrt(37) hat k`Nm

Answer

To find the torque acting on the magnet, we can use the formula for torque \(\vec{\tau}\) in a magnetic field, which is given by:

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

where \(\vec{m}\) is the magnetic moment and \(\vec{B}\) is the magnetic field.

### Step 1: Identify the given values
We have:
- Magnetic moment \(\vec{m} = 50 \hat{i} \, \text{A-m}^2\)
- Magnetic field \(\vec{B} = 0.5 \hat{i} + 3.0 \hat{j} \, \text{T}\)

### Step 2: Write the cross product
We need to calculate the cross product \(\vec{m} \times \vec{B}\):

\[ \vec{\tau} = (50 \hat{i}) \times (0.5 \hat{i} + 3.0 \hat{j}) \]

### Step 3: Distribute the cross product
Using the distributive property of the cross product:

\[ \vec{\tau} = 50 \hat{i} \times (0.5 \hat{i}) + 50 \hat{i} \times (3.0 \hat{j}) \]

### Step 4: Calculate each term
1. For the first term:
\[ 50 \hat{i} \times (0.5 \hat{i}) = 50 \times 0.5 (\hat{i} \times \hat{i}) = 25 \cdot 0 = 0 \]

2. For the second term:
\[ 50 \hat{i} \times (3.0 \hat{j}) = 50 \times 3.0 (\hat{i} \times \hat{j}) = 150 \hat{k} \]

### Step 5: Combine the results
Now, we can combine the results from both terms:

\[ \vec{\tau} = 0 + 150 \hat{k} = 150 \hat{k} \, \text{N-m} \]

### Final Answer
The torque acting on the magnet is:

\[ \vec{\tau} = 150 \hat{k} \, \text{N-m} \]