A current loop, carrying a current of 5.0 A, is in the shape of a right triangle with sides 30,40, and 50 cm. The loop is in a uniform magnetic field of magnitude 80 mT whose direction is parallel to the current in the 50 cm side of the loop. Find the magnitude of (a) the magnetic dipole moment of the loop and (b) the torque on the loop.

Answer

To solve the problem step by step, we will first find the magnetic dipole moment of the current loop and then calculate the torque acting on it.

### Step 1: Calculate the Area of the Triangle
The area \( A \) of a right triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In this case, we can take the two shorter sides (30 cm and 40 cm) as the base and height respectively.

Converting the dimensions from centimeters to meters:
- Base = 30 cm = 0.30 m
- Height = 40 cm = 0.40 m

Now, substituting the values into the area formula:
\[ A = \frac{1}{2} \times 0.30 \, \text{m} \times 0.40 \, \text{m} = \frac{1}{2} \times 0.12 \, \text{m}^2 = 0.06 \, \text{m}^2 \]

### Step 2: Calculate the Magnetic Dipole Moment
The magnetic dipole moment \( \mu \) of a current loop is given by the formula:
\[ \mu = n \cdot I \cdot A \]
where:
- \( n \) = number of turns (which is 1 for a single loop)
- \( I \) = current (5.0 A)
- \( A \) = area calculated in Step 1 (0.06 m²)

Substituting the values:
\[ \mu = 1 \cdot 5.0 \, \text{A} \cdot 0.06 \, \text{m}^2 = 0.30 \, \text{A m}^2 \]

### Step 3: Calculate the Torque on the Loop
The torque \( \tau \) on a current loop in a magnetic field is given by the formula:
\[ \tau = \mu \cdot B \cdot \sin(\theta) \]
where:
- \( B \) = magnetic field strength (80 mT = 80 × 10⁻³ T)
- \( \theta \) = angle between the magnetic moment and the magnetic field (90° in this case, since the direction of the magnetic field is parallel to the current in the 50 cm side)

Since \( \sin(90°) = 1 \), we can simplify the torque formula to:
\[ \tau = \mu \cdot B \]

Substituting the values:
\[ \tau = 0.30 \, \text{A m}^2 \cdot 80 \times 10^{-3} \, \text{T} = 0.30 \cdot 0.08 = 0.024 \, \text{N m} \]

### Final Answers
(a) The magnitude of the magnetic dipole moment of the loop is \( 0.30 \, \text{A m}^2 \).

(b) The torque on the loop is \( 0.024 \, \text{N m} \).