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A cylindrical solenoid that is 40-cm long generates a uniform magnetic field of strength 1.4 T inside the solenoid when the current is 7.0 A. How many turns of a single coil are present in this solenoid?    a. 40,000 turns   b. 64,000 turns   c. 56,000 turns   d. 48,000 turns   e. 32,000 turns

A cylindrical solenoid that is 40-cm long generates a uniform magnetic field of strength 1.4 T inside the solenoid when the current is 7.0 A. How many turns of a single coil are present in this solenoid?    a. 40,000 turns   b. 64,000 turns   c. 56,000 turns   d. 48,000 turns   e. 32,000 turns

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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  1. A cylindrical solenoid that is 40-cm long generates a uniform magnetic field of strength 1.4 T inside the solenoid when the current is 7.0 A. How many turns of a single coil are present in this solenoid? 

      a.

    40,000 turns
      b.

    64,000 turns
      c.

    56,000 turns
      d.

    48,000 turns
      e.

    32,000 turns
### Solenoid Coil Turn Calculation A cylindrical solenoid that is 40 cm long generates a uniform magnetic field of strength 1.4 T inside the solenoid when the current is 7.0 A. How many turns of a single coil are present in this solenoid? - a. 40,000 turns - b. 64,000 turns - c. 56,000 turns - d. 48,000 turns - e. 32,000 turns #### Explanation: This problem involves finding the number of coil turns in a solenoid given its magnetic field strength, length, and current. The number of turns can be calculated using the formula for the magnetic field in a solenoid: \[ B = \mu_0 \times \left( \frac{N}{L} \right) \times I \] Where: - \( B \) is the magnetic field (1.4 T), - \(\mu_0 \) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\)), - \( N \) is the number of turns, - \( L \) is the length of the solenoid (0.4 m), - \( I \) is the current (7.0 A). Solve for \( N \).
Transcribed Image Text:### Solenoid Coil Turn Calculation A cylindrical solenoid that is 40 cm long generates a uniform magnetic field of strength 1.4 T inside the solenoid when the current is 7.0 A. How many turns of a single coil are present in this solenoid? - a. 40,000 turns - b. 64,000 turns - c. 56,000 turns - d. 48,000 turns - e. 32,000 turns #### Explanation: This problem involves finding the number of coil turns in a solenoid given its magnetic field strength, length, and current. The number of turns can be calculated using the formula for the magnetic field in a solenoid: \[ B = \mu_0 \times \left( \frac{N}{L} \right) \times I \] Where: - \( B \) is the magnetic field (1.4 T), - \(\mu_0 \) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\)), - \( N \) is the number of turns, - \( L \) is the length of the solenoid (0.4 m), - \( I \) is the current (7.0 A). Solve for \( N \).
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