Two flat circular coils have a common center but their plane are 90 de...
Given information:
- Inner coil: 150 turns, radius = 1 cm
- Outer coil: 400 turns, radius = 2 cm
- Current through each coil: 200 mA
Calculating the magnetic field produced by each coil:
The magnetic field produced by a current-carrying coil at its center is given by the formula:
B = μ₀ * (N * I) / (2 * R)
Where:
- B is the magnetic field
- μ₀ is the permeability of free space (4π * 10^-7 T·m/A)
- N is the number of turns
- I is the current
- R is the radius of the coil
For the inner coil:
N = 150 turns
I = 200 mA = 0.2 A
R = 1 cm = 0.01 m
Using the formula, we can calculate the magnetic field produced by the inner coil:
B_inner = (4π * 10^-7 T·m/A) * (150 * 0.2 A) / (2 * 0.01 m)
For the outer coil:
N = 400 turns
I = 200 mA = 0.2 A
R = 2 cm = 0.02 m
Using the formula, we can calculate the magnetic field produced by the outer coil:
B_outer = (4π * 10^-7 T·m/A) * (400 * 0.2 A) / (2 * 0.02 m)
Calculating the resultant magnetic field at the center:
Since the two coils are perpendicular to each other, the magnetic fields they produce are at right angles to each other. Thus, the resultant magnetic field can be calculated using the Pythagorean theorem:
B_resultant = √(B_inner² + B_outer²)
Substituting the calculated values, we get:
B_resultant = √(B_inner² + B_outer²)
Final calculation:
Substituting the calculated values of B_inner and B_outer into the equation for B_resultant, we get:
B_resultant = √((4π * 10^-7 T·m/A)² * (150 * 0.2 A)² / (2 * 0.01 m)² + (4π * 10^-7 T·m/A)² * (400 * 0.2 A)² / (2 * 0.02 m)²)
Simplifying the equation gives:
B_resultant ≈ 1.16 * 10^-6 T
Converting the magnetic field to Weber per square meter (Wb/m²):
B_resultant ≈ 1.16 * 10^-6 Wb/m²
Therefore, the correct answer is option (a) 10^-6 Wb/m².