Question

Two solenoids stand side-by-side. Solenoid B differs from solenoid A in three respects. Its total current is 2 times as great, it has 4 times as many coils, and it is 3 times as long. How do the magnetic fields inside the two solenoids compare?

1. Solenoid B’s magnetic field is 24 times greater than solenoid A's magnetic field.
2. Solenoid A’s magnetic field is 24 times greater than solenoid B's magnetic field.
3. Solenoid B’s magnetic field is 6 times greater than solenoid A's magnetic field.
4. Solenoid A’s magnetic field is 2.7 times greater than solenoid B's magnetic field.
5. Solenoid B’s magnetic field is 2.7 times greater than solenoid A's magnetic field.

Answer 1

Solenoid B's magnetic field is 2.7 times greater than the magnetic field of solenoid A because while it has 4 times the number of coils and 2 times the current, it is also 3 times longer, making its turns per unit length ratio and current increase only an overall factor of 2.7.

Step-by-step explanation:

The magnetic field strength inside a solenoid is given by B = μ_0nI, where μ_0 is the magnetic constant (permeability of free space), n is the number of turns per unit length, and I is the current. For solenoid B, since it has a current 2 times greater, 4 times as many coils, and is 3 times longer than solenoid A, we need to calculate how these changes affect the magnetic field.

The number of turns per unit length for solenoid B is 4 times greater than that of solenoid A, but since it's also 3 times longer, the number of turns per unit length (n) is effectively 4/3 times that of solenoid A. Multiplying this by the 2 times greater current gives us an overall factor of 8/3 times the magnetic field of solenoid A. Therefore, Solenoid B's magnetic field is 2.7 times greater than solenoid A's magnetic field.

Answer 2

5. Solenoid B’s magnetic field is 2.7 times greater than solenoid A's magnetic field.

Step-by-step explanation:

The magnetic field produced by a solenoid depends on number of coils, the length of the solenoid and the amount of current flowing through the coil.

`B=(\mu * N * I)/(L)`

Current in solenoid B is 2 times greater than the current in solenoid A.

`I_B=2I_A`

Number of coils in B is 4 times greater than coils in A

`N_B=4N_A`

Length of the solenoid B is 3 times greater than solenoid A

`L_B=3L_A`

`B_B=(2* 4)/(3) B_A=2.7 B_A`

Thus, solenoid B's magnetic field is 2.7 times greater than magnetic field of solenoid A.