A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that `vecB` is in a plane of the coil. If due ot a current I, in the triangle a torque `tau` acts on it the side l of the triangle is given by `N((tau)/(BiP))^(1/2)`. What will be the numerical value of `N+P`?

Explanation

To solve the problem, we need to derive the expression for the side length \( l \) of the equilateral triangle coil in terms of the given quantities and identify the constants \( N \) and \( P \).

### Step 1: Understanding the Torque on the Coil
The torque \( \tau \) acting on a current-carrying loop in a magnetic field is given by the formula:
\[ \tau = N \cdot I \cdot A \cdot B \cdot \sin(\theta) \]
where:
- \( N \) is the number of turns (which is 1 for a single triangular coil),
- \( I \) is the current flowing through the coil,
- \( A \) is the area of the coil,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the area vector and the magnetic field.

### Step 2: Area of the Equilateral Triangle
For an equilateral triangle with side length \( l \), the area \( A \) can be calculated using the formula:
\[ A = \frac{\sqrt{3}}{4} l^2 \]

### Step 3: Angle Between Area Vector and Magnetic Field
Since the magnetic field \( \vec{B} \) is in the same plane as the coil, the angle \( \theta \) between the area vector (which is perpendicular to the plane of the triangle) and the magnetic field is \( 90^\circ \). Thus, \( \sin(90^\circ) = 1 \).

### Step 4: Substituting Values into the Torque Equation
Substituting the values into the torque equation:
\[ \tau = 1 \cdot I \cdot \left(\frac{\sqrt{3}}{4} l^2\right) \cdot B \cdot 1 \]
This simplifies to:
\[ \tau = \frac{\sqrt{3}}{4} I B l^2 \]

### Step 5: Rearranging for Side Length \( l \)
To find the side length \( l \), we rearrange the equation:
\[ l^2 = \frac{4\tau}{\sqrt{3} I B} \]
Taking the square root gives:
\[ l = \sqrt{\frac{4\tau}{\sqrt{3} I B}} = \frac{2\sqrt{\tau}}{(I B)^{1/2} \cdot \sqrt[4]{3}} \]

### Step 6: Identifying Constants \( N \) and \( P \)
From the expression given in the problem:
\[ l = N \left(\frac{\tau}{B I P}\right)^{1/2} \]
We can compare this with our derived expression:
\[ l = \frac{2\sqrt{\tau}}{(I B)^{1/2} \cdot \sqrt[4]{3}} \]
From this, we can identify:
- \( N = 2 \)
- \( P = \sqrt{3} \)

### Step 7: Calculating \( N + P \)
Now, we need to find the numerical value of \( N + P \):
\[ N + P = 2 + \sqrt{3} \]
The approximate value of \( \sqrt{3} \) is \( 1.732 \), thus:
\[ N + P \approx 2 + 1.732 = 3.732 \]

### Final Answer
The numerical value of \( N + P \) is approximately \( 3.732 \).

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