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The length of a solenoil is 0.1 metre and diameter is very small a wire is wound over it is two layers. The number of turns in the inner layer is 50 and that on outer layer is 40 the strength of current falling in two layers is in the same direction and is 3 ampere. The magnetic induction in the middle of the solenoid will be-?
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The length of a solenoil is 0.1 metre and diameter is very small a wir...
Calculating the Magnetic Induction in the Middle of the Solenoid
- **Given Parameters**:
- Length of solenoid (l) = 0.1 m
- Number of turns in inner layer (N1) = 50
- Number of turns in outer layer (N2) = 40
- Current in both layers (I) = 3 A
- **Formula for Magnetic Induction (B) in the middle of a Solenoid**:
- B = μ₀ * n * I
- Where:
- μ₀ = Permeability of free space (4π x 10^-7 T m/A)
- n = Total number of turns per unit length
- **Calculating Total Number of Turns per Unit Length (n)**:
- Total turns in both layers = N1 + N2 = 50 + 40 = 90 turns
- Total length of wire in two layers = 2 * l = 2 * 0.1 m = 0.2 m
- Total number of turns per unit length (n) = Total turns / Total length = 90 / 0.2 = 450 turns/m
- **Calculating Magnetic Induction (B) in the Middle of the Solenoid**:
- B = μ₀ * n * I
- B = 4π x 10^-7 T m/A * 450 turns/m * 3 A
- B = 1.8 x 10^-4 T
Therefore, the magnetic induction in the middle of the solenoid is 1.8 x 10^-4 Tesla. This value represents the strength of the magnetic field inside the solenoid due to the current flowing through the wire wound around it.
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The length of a solenoil is 0.1 metre and diameter is very small a wire is wound over it is two layers. The number of turns in the inner layer is 50 and that on outer layer is 40 the strength of current falling in two layers is in the same direction and is 3 ampere. The magnetic induction in the middle of the solenoid will be-?
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