A solenoid 1.30 m long and 2.60 cm in diameter carries a current of 18.0 A. The magnitude of

step 1 Here we have given the magnetic field inside the solenide that is equal to our 23 micro -Tesla,

A solenoid 1.30 m long and 2.60 cm in diameter carries a...

A solenoid $1.30 \mathrm{~m}$ long and $2.60 \mathrm{~cm}$ in diameter carries a current of $18.0 \mathrm{~A}$. The magnitude of the magnetic field inside the solenoid is $23.0 \mathrm{mT}$. Find the length of the wire forming the solenoid.

Key Concepts

Magnetic Field in a Solenoid

This concept explains how the magnetic field inside an ideal solenoid is produced by the flow of electric current through closely spaced coils. The field is uniformly strong and aligned along the solenoid’s axis, and its magnitude depends on the current, the number of turns per unit length, and the permeability of free space as expressed by the formula B = ?? (N/L) I.

Turn Density and Number of Turns

Turn density refers to the number of windings per unit length of the solenoid and is a crucial factor in determining the magnetic field strength. By knowing the desired field and the current, one can calculate the total number of turns required by rearranging the relation B = ?? (N/L) I to find N, the total number of turns.

Circumference of a Coil Turn

Each loop or turn of the solenoid is essentially a circle, and the length of wire used in each turn is given by its circumference. Knowing the diameter (or radius) of the solenoid allows the calculation of this individual loop length, which is then used to determine the total wire length when multiplied by the total number of turns.

Total Wire Length Calculation

The total length of wire forming the solenoid is calculated by summing the lengths of all individual turns. This is achieved by multiplying the number of turns by the circumference of one turn, linking the magnetic field requirements with the physical dimensions of the solenoid.

Transcript

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