The plane of a rectangular loop of wire with sides 0.05 m and 0.08 m is parallel to a uniform magnetic field of induction `1.5 xx 10^-2 T.` A current of 10.0 ampere flows through the loop. If the side of length 0.08 m is normal and the side of length 0.05 m is parallel to the lines of induction, then the torque acting on the loop is

Options

A. `6000 Nm`
B. zero
C. `1.2 xx 10^-2 Nm`
D. `6 xx 10^-4 Nm`

Explanation

To solve the problem, we need to calculate the torque acting on a rectangular loop of wire in a magnetic field. The torque (\( \tau \)) on a current-carrying loop in a magnetic field is given by the formula:

\[ \tau = m \times B \times \sin(\theta) \]

where:
- \( m \) is the magnetic moment of the loop,
- \( B \) is the magnetic field induction,
- \( \theta \) is the angle between the magnetic moment and the magnetic field.

### Step 1: Calculate the area of the loop
The area \( A \) of the rectangular loop can be calculated using the formula:

\[ A = \text{length} \times \text{width} \]

Given that the sides of the loop are 0.05 m and 0.08 m:

\[ A = 0.05 \, \text{m} \times 0.08 \, \text{m} = 0.004 \, \text{m}^2 \]

### Step 2: Calculate the magnetic moment \( m \)
The magnetic moment \( m \) is given by the product of the current \( I \) flowing through the loop and the area \( A \):

\[ m = I \times A \]

Given that the current \( I = 10.0 \, \text{A} \):

\[ m = 10.0 \, \text{A} \times 0.004 \, \text{m}^2 = 0.04 \, \text{A} \cdot \text{m}^2 \]

### Step 3: Determine the angle \( \theta \)
In this case, the side of length 0.08 m is normal to the magnetic field, and the side of length 0.05 m is parallel to the magnetic field lines. Therefore, the angle \( \theta \) between the magnetic moment and the magnetic field is \( 90^\circ \).

\[ \sin(\theta) = \sin(90^\circ) = 1 \]

### Step 4: Calculate the torque \( \tau \)
Now we can substitute the values into the torque formula:

\[ \tau = m \times B \times \sin(\theta) \]

Given that the magnetic field induction \( B = 1.5 \times 10^{-2} \, \text{T} \):

\[ \tau = 0.04 \, \text{A} \cdot \text{m}^2 \times 1.5 \times 10^{-2} \, \text{T} \times 1 \]

Calculating this gives:

\[ \tau = 0.04 \times 1.5 \times 10^{-2} = 0.0006 \, \text{N} \cdot \text{m} = 6 \times 10^{-4} \, \text{N} \cdot \text{m} \]

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

\[ \tau = 6 \times 10^{-4} \, \text{N} \cdot \text{m} \]