A dipole of magnetic moment `vec(m)=30hatjA m^(2)` is placed along the y-axis in a uniform magnetic field `vec(B)= (2hat i +5hatj)T`. The torque acting on it is

Options

A. `-40hatk Nm`
B. `-50hatk Nm`
C. `-60hatk Nm`
D. `-70hatk Nm`

Explanation

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

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

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnetic moment: \(\vec{m} = 30 \hat{j} \, \text{A m}^2\)
- Magnetic field: \(\vec{B} = 2 \hat{i} + 5 \hat{j} \, \text{T}\)

2. **Set Up the Torque Equation:**
- Using the formula for torque:
\[ \vec{\tau} = \vec{m} \times \vec{B} \]

3. **Substitute the Values:**
- Substitute the values of \(\vec{m}\) and \(\vec{B}\):
\[ \vec{\tau} = (30 \hat{j}) \times (2 \hat{i} + 5 \hat{j}) \]

4. **Distribute the Cross Product:**
- Use the distributive property of the cross product:
\[ \vec{\tau} = 30 \hat{j} \times (2 \hat{i}) + 30 \hat{j} \times (5 \hat{j}) \]

5. **Calculate Each Component:**
- For the first term:
\[ 30 \hat{j} \times (2 \hat{i}) = 60 (\hat{j} \times \hat{i}) = 60 (-\hat{k}) = -60 \hat{k} \]
- For the second term:
\[ 30 \hat{j} \times (5 \hat{j}) = 150 (\hat{j} \times \hat{j}) = 150 (0) = 0 \]

6. **Combine the Results:**
- Combine the results from both terms:
\[ \vec{\tau} = -60 \hat{k} + 0 = -60 \hat{k} \, \text{N m} \]

7. **Final Result:**
- The torque acting on the dipole is:
\[ \vec{\tau} = -60 \hat{k} \, \text{N m} \]