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) N-m`
B. `150 hat(k) N-m`
C. `75 hat(k) N-m`
D. `25sqrt(37) hat(k) N-m`

Explanation

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

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

where:
- \(\vec{m}\) is the magnetic moment,
- \(\vec{B}\) is the magnetic field,
- \(\times\) denotes the cross product.

### Step 1: Identify the given values
The magnetic moment is given as:
\[ \vec{m} = 50 \hat{i} \, \text{A-m}^2 \]
The magnetic field is given as:
\[ \vec{B} = 0.5 \hat{i} + 3.0 \hat{j} \, \text{T} \]

### Step 2: Set up the cross product
Now, we need to calculate the cross product \(\vec{m} \times \vec{B}\):
\[ \tau = \vec{m} \times \vec{B} = (50 \hat{i}) \times (0.5 \hat{i} + 3.0 \hat{j}) \]

### Step 3: Distribute the cross product
Using the distributive property of the cross product:
\[ \tau = (50 \hat{i}) \times (0.5 \hat{i}) + (50 \hat{i}) \times (3.0 \hat{j}) \]

### Step 4: Calculate each term
1. The first term:
\[ (50 \hat{i}) \times (0.5 \hat{i}) = 50 \times 0.5 (\hat{i} \times \hat{i}) = 25 \times 0 = 0 \]
(since the cross product of any vector with itself is zero)

2. The second term:
\[ (50 \hat{i}) \times (3.0 \hat{j}) = 50 \times 3.0 (\hat{i} \times \hat{j}) = 150 \hat{k} \]
(using the right-hand rule, \(\hat{i} \times \hat{j} = \hat{k}\))

### Step 5: Combine the results
Combining both results, we have:
\[ \tau = 0 + 150 \hat{k} = 150 \hat{k} \, \text{N-m} \]

### Final Answer
The torque acting on the magnet is:
\[ \tau = 150 \hat{k} \, \text{N-m} \]