A planar coil having 15turns carries 20 A currennt . The coil is oriented with respect to the unifrom magnetic field B = 0.5 T such that its direction area is `A = - 0.04hatim^(2)`. The potenial energy of the coil in the given orientation is

Options

A. 0
B. `+ 0.72`
C. 6J
D. `-1.44 J`

Explanation

To find the potential energy of the coil in the given orientation, we can follow these steps:

### Step 1: Determine the Magnetic Moment (m)
The magnetic moment \( m \) of a coil is given by the formula:
\[ m = n \cdot I \cdot A \]
where:
- \( n \) = number of turns (15 turns)
- \( I \) = current (20 A)
- \( A \) = area of the coil

Given that the area vector \( A = -0.04 \hat{i} \, m^2 \), we can calculate the magnitude of the area:
\[ |A| = 0.04 \, m^2 \]

Now substituting the values:
\[ m = 15 \cdot 20 \cdot 0.04 = 12 \, A \cdot m^2 \]
Thus, the magnetic moment vector is:
\[ m = 12 \hat{i} \, A \cdot m^2 \]

### Step 2: Identify the Magnetic Field (B)
The magnetic field is given as:
\[ B = 0.5 \hat{i} \, T \]

### Step 3: Calculate the Angle (θ)
The area vector \( A \) is oriented in the negative direction of the x-axis, while the magnetic field \( B \) is in the positive direction of the x-axis. Therefore, the angle \( \theta \) between the magnetic moment and the magnetic field is:
\[ \theta = 180^\circ = \pi \, \text{radians} \]

### Step 4: Calculate the Potential Energy (U)
The potential energy \( U \) of the coil in a magnetic field is given by:
\[ U = -\vec{m} \cdot \vec{B} = -mB \cos(\theta) \]

Substituting the values:
\[ U = - (12 \, A \cdot m^2)(0.5 \, T) \cos(\pi) \]
Since \( \cos(\pi) = -1 \):
\[ U = - (12)(0.5)(-1) = 6 \, J \]

### Final Answer
The potential energy of the coil in the given orientation is:
\[ U = 6 \, J \]

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