What is magnetic flux linked with a coil of N turns and cross section area A held with its plane parallel to the field?

Options

A. `(NAB)/(2)`
B. NAB
C. `(NAB)/(4)`
D. zero

Explanation

To find the magnetic flux linked with a coil of N turns and cross-sectional area A held with its plane parallel to the magnetic field, we can follow these steps:

### Step 1: Understand the Concept of Magnetic Flux
Magnetic flux (Φ) through a surface is defined as the product of the magnetic field (B) and the area (A) of the surface, taking into account the angle (θ) between the magnetic field lines and the normal (perpendicular) to the surface.

### Step 2: Write the Formula for Magnetic Flux
The formula for magnetic flux linked with a coil is given by:
\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]
where:
- \( \Phi \) = magnetic flux
- \( N \) = number of turns in the coil
- \( B \) = magnitude of the magnetic field
- \( A \) = cross-sectional area of the coil
- \( \theta \) = angle between the normal to the coil's plane and the magnetic field direction

### Step 3: Determine the Angle θ
In this case, the coil is held with its plane parallel to the magnetic field. This means that the angle θ between the normal to the coil's plane and the direction of the magnetic field is 90 degrees (θ = 90°).

### Step 4: Substitute θ into the Formula
Substituting θ = 90° into the formula, we have:
\[ \Phi = N \cdot B \cdot A \cdot \cos(90°) \]

### Step 5: Calculate cos(90°)
We know that:
\[ \cos(90°) = 0 \]

### Step 6: Final Calculation
Substituting cos(90°) = 0 into the equation:
\[ \Phi = N \cdot B \cdot A \cdot 0 = 0 \]

### Conclusion
Thus, the magnetic flux linked with the coil held with its plane parallel to the magnetic field is:
\[ \Phi = 0 \]

### Answer
The magnetic flux linked with the coil is 0.

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