A coil in the shape of an equilateral triangle of side 0.02 m is suspended from a vertex such that it is hanging in a vertical plane between the pole pieces of a permanent magnet producing a horizontal magnetic field of `5 xx 10^(-2) T.` Find the couple acting on the coil when a current of 0.1 A is passed through it and the magnetic field is parallel to its plane.

Options

A. `5sqrt(3)xx10^(-7)N-m`
B. `15sqrt(3)xx10^(-10)N-m`
C. `(sqrt(3))/(15)xx10^(-7)N-m`
D. none of these

Answer

To find the couple (torque) acting on the coil when a current is passed through it, we can follow these steps:

### Step 1: Understand the Problem
We have a coil in the shape of an equilateral triangle suspended from a vertex in a vertical plane. It is placed in a horizontal magnetic field of strength \( B = 5 \times 10^{-2} \, \text{T} \). The current flowing through the coil is \( I = 0.1 \, \text{A} \).

### Step 2: Calculate the Area of the Triangle
The area \( A \) of an equilateral triangle can be calculated using the formula:
\[ A = \frac{\sqrt{3}}{4} a^2 \]
where \( a \) is the side length of the triangle. Given \( a = 0.02 \, \text{m} \):
\[ A = \frac{\sqrt{3}}{4} (0.02)^2 = \frac{\sqrt{3}}{4} \times 0.0004 = \frac{\sqrt{3}}{16000} \, \text{m}^2 \]

### Step 3: Calculate the Magnetic Moment
The magnetic moment \( \mu \) of the coil is given by:
\[ \mu = nIA \]
where \( n \) is the number of turns (which is 1 for this problem), \( I \) is the current, and \( A \) is the area calculated above. Thus:
\[ \mu = 1 \times 0.1 \times A = 0.1 \times \frac{\sqrt{3}}{16000} = \frac{0.1\sqrt{3}}{16000} \, \text{A m}^2 \]

### Step 4: Calculate the Torque
The torque \( \tau \) acting on the coil in a magnetic field is given by:
\[ \tau = \mu B \sin \theta \]
In this case, since the magnetic field is parallel to the plane of the coil, \( \theta = 90^\circ \) and \( \sin 90^\circ = 1 \):
\[ \tau = \mu B \]
Substituting the values:
\[ \tau = \left(\frac{0.1\sqrt{3}}{16000}\right) \times (5 \times 10^{-2}) \]

### Step 5: Simplify the Expression
Calculating the torque:
\[ \tau = \frac{0.1 \times 5 \times 10^{-2} \times \sqrt{3}}{16000} = \frac{0.005\sqrt{3}}{16000} \]
\[ \tau = \frac{5\sqrt{3}}{1600000} = \frac{\sqrt{3}}{320000} \]

### Step 6: Final Calculation
Now we can approximate the value of \( \sqrt{3} \approx 1.732 \):
\[ \tau \approx \frac{1.732}{320000} \approx 5.4 \times 10^{-6} \, \text{N m} \]

### Conclusion
The couple acting on the coil is approximately \( 5.4 \times 10^{-6} \, \text{N m} \).

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