Understanding Magnetic Field in a Solenoid

This question asks about the magnetic field strength inside a solenoid and how it changes when the number of turns per unit length is altered.

A solenoid is essentially a coil of wire wound into a tightly packed helix. When an electric current flows through the wire of a solenoid, it produces a magnetic field. The magnetic field inside a long solenoid is known to be very uniform and strong, particularly along its axis, and relatively weak outside.

Formula for Magnetic Field Inside a Solenoid

The magnitude of the magnetic field (\(B\)) inside a long solenoid is given by the formula:

\(B = \mu_0 n I\)

Where:

• \(B\) is the magnitude of the magnetic field inside the solenoid.
• \(\mu_0\) is the permeability of free space, a constant value.
• \(n\) is the number of turns per unit length of the solenoid.
• \(I\) is the current flowing through the wire of the solenoid.

This formula shows that the magnetic field inside the solenoid is directly proportional to the permeability of the medium (\(\mu_0\)), the number of turns per unit length (\(n\)), and the current (\(I\)) flowing through the wire.

Analyzing the Initial and Changed Conditions

Let's break down the given information:

Initial State:

• Current flowing through the wire = \(I\)
• Number of turns per unit length = \(n\)
• Magnetic field inside the solenoid = \(B\)

Using the formula, the initial magnetic field is \(B = \mu_0 n I\).

Changed State:

• Current flowing through the wire remains = \(I\) (The question states the current is \(I\)).
• Number of turns per unit length is increased to = \(2n\)
• Magnetic field inside the solenoid = Let's call this new field \(B'\).

Now, let's calculate the new magnetic field \(B'\) using the same formula but with the new number of turns per unit length (\(2n\)) and the same current (\(I\)).

Calculating the New Magnetic Field

For the changed state, the formula for the magnetic field \(B'\) is:

\(B' = \mu_0 (2n) I\)

\(B' = 2 \mu_0 n I\)

We know from the initial state that the original magnetic field \(B\) was given by \(B = \mu_0 n I\).

Comparing the expression for \(B'\) with the expression for \(B\), we can see that:

\(B' = 2 (\mu_0 n I)\)

\(B' = 2B\)

Therefore, when the number of turns per unit length is doubled (from \(n\) to \(2n\)) while the current remains the same, the magnetic field inside the solenoid doubles.

Comparing Results

Initial magnetic field = \(B\)

New magnetic field = \(B'\)

We found that \(B' = 2B\).

This means the new magnetic field is twice the original magnetic field.

Let's look at the options provided:

1. B
2. 2B
3. B / 2
4. B / 4

Our calculated new magnetic field is \(2B\), which matches option 2.

Conclusion on Magnetic Field Change

The magnetic field inside a long solenoid is directly proportional to the number of turns per unit length (\(n\)) and the current (\(I\)). When \(n\) is doubled and \(I\) remains constant, the magnetic field strength also doubles. This is a direct consequence of the linear relationship between \(B\), \(n\), and \(I\) in the formula \(B = \mu_0 n I\).

Revision Table: Solenoid Properties and Magnetic Field

Additional Information: Long vs. Finite Solenoids

The formula \(B = \mu_0 n I\) is strictly accurate for an infinitely long solenoid or approximately accurate for a very long solenoid far from its ends. For a finite length solenoid, the magnetic field is strongest and most uniform near the center but decreases as you move towards the ends. At the ends of a long solenoid, the magnetic field strength is approximately half of the value at the center (\(B_{end} \approx \frac{1}{2} \mu_0 n I\)). However, for the purpose of this type of problem involving a solenoid, it is generally assumed to be long enough for the uniform field formula to apply unless otherwise specified.

Also, the magnetic field inside a solenoid is parallel to its axis.