A dipole of magnetic moment `vec(m)=(30 f) (A) (m)^(2)` is placed along the y-axis in a, uniform magnetic field `vec(B)=(2 hat(i)+5 hat(j)) . (T)`. The torque acting 'on it is `(-alpha hat(k))`. Calculate `alpha`.

Answer

To solve the problem, we need to calculate the torque acting on a magnetic dipole in a uniform magnetic field using the formula for torque, which is given by:

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

### Step 1: Identify the magnetic moment and magnetic field vectors

The magnetic moment \(\vec{m}\) is given as:

\[ \vec{m} = 30 \hat{f} \, \text{A m}^2 \]

Since it is placed along the y-axis, we can express it as:

\[ \vec{m} = 30 \hat{j} \, \text{A m}^2 \]

The magnetic field \(\vec{B}\) is given as:

\[ \vec{B} = 2 \hat{i} + 5 \hat{j} \, \text{T} \]

### Step 2: Calculate the cross product \(\vec{m} \times \vec{B}\)

Using the formula for the cross product, we have:

\[ \vec{\tau} = \vec{m} \times \vec{B} = (30 \hat{j}) \times (2 \hat{i} + 5 \hat{j}) \]

### Step 3: Expand the cross product

Using the distributive property of the cross product:

\[ \vec{\tau} = 30 \hat{j} \times (2 \hat{i}) + 30 \hat{j} \times (5 \hat{j}) \]

Since \(\hat{j} \times \hat{j} = 0\), the second term vanishes:

\[ \vec{\tau} = 30 \hat{j} \times (2 \hat{i}) = 60 (\hat{j} \times \hat{i}) \]

### Step 4: Determine the direction of \(\hat{j} \times \hat{i}\)

Using the right-hand rule, we find that:

\[ \hat{j} \times \hat{i} = -\hat{k} \]

Thus, we have:

\[ \vec{\tau} = 60 (-\hat{k}) = -60 \hat{k} \]

### Step 5: Compare with the given torque

The problem states that the torque acting on it is \(-\alpha \hat{k}\). From our calculation, we have:

\[ \vec{\tau} = -60 \hat{k} \]

This implies:

\[ -\alpha = -60 \]

### Step 6: Solve for \(\alpha\)

By comparing both sides, we find:

\[ \alpha = 60 \]

### Final Answer

Thus, the value of \(\alpha\) is:

\[ \alpha = 60 \]

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