Question

A long solenoid (1 000 turns/m) carries a current of 25 mA and has an inside radius of 2.0 cm. A long wire which is parallel to and 4.0 cm from the axis of the solenoid carries a current of 6.0 A. What is the magnitude of the magnetic field at a point on the axis of the solenoid

Answer 1

Answer:

B=4.343 * 10⁻⁵ T

Explanation:

See attached photo for explanation.

Answer 2

The magnitude of the magnetic field at a point on the axis of the solenoid is 3 x 10⁻⁵ T.

The given parameters;

• number of turns per length = 1000 turns/m
• current in the solenoid, = 25 mA = 0.025 A
• distance from the axis of the solenoid, r = 4 cm
• current at the axis of the solenoid I, = 6 A

The magnitude of the magnetic field at a point on the axis of the solenoid is calculated as follows;

$B = \frac{\mu_0 I}{2\pi r}$

where;

• $\mu_o$ is permittivity of free space = 4π x 10⁻⁷ N/A²
• B is the magnitude of the magnetic field

The value of the magnetic field is calculated by substituting the parameters;

$B = \frac{(4\pi \times 10^{-7}) \times 6}{2\pi (0.04)} \\\\B = 3\times 10^{-5} \ T$

Thus, the magnitude of the magnetic field at a point on the axis of the solenoid is 3 x 10⁻⁵ T.

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