The torque acting on a magnetie dipole of moment `5 A m^(2)` when placed in an external uniform magnetic induction `1.5 xx 10^(-4)" Wb/m"^(2)` at right angle to magnetic induction is

Options

A. `7.5 xx 10^(-4) Nm`
B. `75 xx 10^(-4) Nm `
C. `1.25 xx 10^(-5) Nm`
D. `1.5 xx 10^(-4) Nm`

Answer

To find the torque acting on a magnetic dipole when placed in an external uniform magnetic field, we can follow these steps:

### Step 1: Understand the Given Values
We are given:
- Magnetic dipole moment (m) = 5 A·m²
- Magnetic field induction (B) = 1.5 × 10^(-4) Wb/m² (which is equivalent to Tesla)
- The angle (θ) between the magnetic dipole moment and the magnetic field is 90 degrees (perpendicular).

### Step 2: Use the Formula for Torque
The torque (τ) acting on a magnetic dipole in a magnetic field is given by the formula:
\[ \tau = m \times B \times \sin(\theta) \]
Since the dipole is placed at a right angle to the magnetic field, we have:
\[ \sin(90^\circ) = 1 \]
Thus, the formula simplifies to:
\[ \tau = m \times B \]

### Step 3: Substitute the Values
Now, substitute the values of m and B into the formula:
\[ \tau = 5 \, \text{A·m}^2 \times 1.5 \times 10^{-4} \, \text{T} \]

### Step 4: Calculate the Torque
Now, perform the multiplication:
\[ \tau = 5 \times 1.5 \times 10^{-4} \]
\[ \tau = 7.5 \times 10^{-4} \, \text{N·m} \]

### Step 5: Conclusion
The torque acting on the magnetic dipole is:
\[ \tau = 7.5 \times 10^{-4} \, \text{N·m} \]

### Final Answer
The torque acting on the magnetic dipole is \( 7.5 \times 10^{-4} \, \text{N·m} \).

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