A linear solenoid is made of a copper wire of diameter 0.4 mm. Its length is 10 cm, diameter 1 cm and number of turns 1000. If the two ends of the solenoid is connected to a battery of emf 2 V, determine the magnetic field produced at the mid-point on the axis of the solenoid. Resistivity of copper = 1.76 x 10-82 ohm m.
ANSWER- [5.7 x 10^(-3) T]
Answers
Answer:
5.7 × 10⁻³ T
Explanation:
Length of the copper wire is
L = 2πrN (where N is number of turns)
= 2 × π × 0.5 cm × 1000
= 1000π cm
= 10π m
Resistance in the wire is
R = ρL / A
= (1.76 × 10⁻⁸ Ω m × 10π m) / [π (0.2 × 10⁻³ m)²]
= 1.76 × 10⁻⁷ / (4 × 10⁻⁸) Ω
= 4.4 Ω
Current in the solenoid = V/R
= 2 V / (4.4 Ω)
= 5/11 A
Magnetic field produced is
- B = μnI
Where
- μ = Permeability
- n = Turns density (N/L)
- I = Current in the solenoid
B = 4π × 10⁻⁷ × (1000 / 0.1 m) × 5/11 A
= 4 × 22/7 × 5/11 × 10⁻⁷⁺⁴ T
= 5.7 × 10⁻³ T
Answer:
Length of the copper wire is
L = 2πrN (where N is number of turns)
= 2 × π × 0.5 cm × 1000
= 1000π cm
= 10π m
Resistance in the wire is
R = ρL / A
= (1.76 × 10⁻⁸ Ω m × 10π m) / [π (0.2 × 10⁻³ m)²]
= 1.76 × 10⁻⁷ / (4 × 10⁻⁸) Ω
= 4.4 Ω
Current in the solenoid = V/R
= 2 V / (4.4 Ω)
= 5/11 A
Magnetic field produced is
B = μnI
Where
μ = Permeability
n = Turns density (N/L)
I = Current in the solenoid
B = 4π × 10⁻⁷ × (1000 / 0.1 m) × 5/11 A
= 4 × 22/7 × 5/11 × 10⁻⁷⁺⁴ T
= 5.7 × 10⁻³ T
Explanation:
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