Problem 23 - Master Resource Book in JEE Main Physics

Question

A closely wound solenoid \(80 \mathrm{~cm}\) long has 5 layers of windings of 400 turns each. The diameter of the solenoid is \(1.8 \mathrm{~cm} .\) If the current carried is \(8.0 \mathrm{~A}\), estimate the magnitude of \(B\) inside the solenoid near its centre. \(\quad\) (a) \(1.5 \times 10^{-2} \mathrm{~T}\), opposite to the axis of solenoid (b) \(2.5 \times 10^{-2} \mathrm{~T}\), along the axis of solenoid (c) \(3.5 \times 10^{-2} \mathrm{~T}\), along the axis of solenoid (d) \(1.5 \times 10^{-2} \mathrm{~T}\), opposite to the axis of solenoid

Step-by-Step Solution

Answer

The magnitude of \(B\) is (b) \(2.5 \times 10^{-2} \mathrm{~T}\), along the axis of solenoid.

Step 1: Understand the Problem

We need to determine the magnetic field, denoted as \(B\), inside a solenoid given the number of turns, the length of the solenoid, and the current flowing through it. We are given that the solenoid is closely wound, which means we can assume the field inside is uniform and use the formula for the magnetic field in a solenoid.

Step 2: Recall the Formula for B Inside a Solenoid

The magnetic field \(B\) inside a long tightly wound solenoid with \(n\) turns per unit length can be calculated using the formula: \[ B = \mu_0 n I \] where \(\mu_0\) is the permeability of free space \((4\pi \times 10^{-7} \mathrm{~Tm/A})\), \(n\) is the number of turns per unit length, and \(I\) is the current.

Step 3: Calculate the Total Number of Turns

The solenoid has 5 layers with 400 turns each. So, the total number of turns \(N\) is \(5 \times 400 = 2000\) turns.

Step 4: Determine the Number of Turns per unit Length (n)

The length \(L\) of the solenoid is \(80\) cm, which is \(0.8\) m. The number of turns per unit length \(n\) is \[ n = \frac{N}{L} = \frac{2000}{0.8} \approx 2500 \, \text{turns/m} \]

Step 5: Substitute Values into the B Formula

Now we substitute \(\mu_0 = 4\pi \times 10^{-7}\), \(n = 2500\), and \(I = 8\) A into the formula: \[ B = \mu_0 n I = (4\pi \times 10^{-7}) \times 2500 \times 8 \approx 2.51 \times 10^{-2} \, \text{T} \]

Step 6: Determine the Direction of B

Since the solenoid carries current, the direction of \(B\) is along the axis of the solenoid, following the right-hand rule. The magnetic field lines inside the solenoid run parallel to the axis.

Step 7: Choose the Correct Option

The calculated magnitude is \(2.51 \times 10^{-2} \, \text{T}\), and the direction is along the axis of the solenoid. Therefore, the correct answer is (b) \(2.5 \times 10^{-2} \, \text{T}\), along the axis of solenoid.

Key Concepts

Permeability of Free Space

The permeability of free space, denoted as \(\mu_0\), is a fundamental physical constant used in electromagnetism. It reflects the ability of a vacuum to support the formation of a magnetic field and is integral to the calculation of magnetic fields in various contexts, including solenoids.
\[\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}\]
This constant links magnetic fields to the currents that produce them, playing a pivotal role in calculations. When a solenoid is involved, \(\mu_0\) helps determine the magnitude of the magnetic field within it by utilizing the formula for the magnetic field in a solenoid, often expressed as \(B = \mu_0\cdot n\cdot I\) where \(n\) is the number of turns per unit length and \(I\) is the current.

• \(\mu_0\) is crucial for formulating equations related to magnetic fields.
• It ensures standardization in scientific calculations involving magnetism.
• Understanding its role helps in grasping how magnetic fields behave in a vacuum.

Right-Hand Rule

The right-hand rule is a simple mnemonic used to determine the direction of the magnetic field relative to the direction of the current. To apply it, point your right thumb in the direction of the electric current through the solenoid, and curl your fingers.
The direction in which your fingers curl shows the direction of the magnetic field lines inside the solenoid.
This establishes that the field lines within a solenoid are parallel to its axis when the wire carries current.

• It is essential for visualizing the orientation of the field relative to the current.
• Keeps the understanding of magnetic fields consistent and easy to follow.
• Assists in practical applications, such as determining the direction of force on a charged particle moving in a magnetic field.

Understanding the right-hand rule clarifies the behavior of magnetic fields and current in conductors and solenoids, a common topic in physics.

Current in Solenoid

In the context of a solenoid, current is the flow of electric charge through the wire coils. This current generates a magnetic field inside the solenoid.
The magnitude of the current, denoted as \(I\), is a critical factor influencing the strength of this magnetic field. As the amount of current increases, the magnetic field strength also increases, following a directly proportional relationship.
For instance, in the given exercise, a current of 8.0 A contributes to the magnetic field inside the solenoid. The interaction between current and the solenoid ensures that the magnetic field is strong and concentrated along the solenoid's axis.

• Current flow is necessary for inducing magnetic fields within solenoids.
• It is directly proportional to the magnetic field strength. Higher current results in a stronger field.
• Understanding this relationship is vital for applications requiring controlled magnetic fields, such as electromagnets and inductors.

Magnetic Field Calculation

Calculating the magnetic field inside a solenoid is straightforward if you understand the relevant formula: \(B = \mu_0 \cdot n \cdot I\). This formula calculates the magnetic flux density, \(B\), and depends on three variables:
- **\(\mu_0\)**: The permeability of free space, which is constant.
- **\(n\)**: The number of turns per unit length, calculated as total turns divided by the length of the solenoid.
- **\(I\)**: The current flowing through the solenoid.
To find \(n\), divide the total number of turns by the solenoid's length. Substitute all known values into the formula for \(B\), ensuring unit consistency.
In this particular exercise, substituting \(\mu_0 = 4\pi \times 10^{-7}\), \(n = 2500\), and \(I = 8 \text{ A}\) gives \(B = 2.5 \times 10^{-2} \text{ T}\), indicating the field's magnitude and direction along the solenoid's axis.

• Accurate calculation requires correct use of constants and conversion of measurements.
• Understanding the relationship between the formula's components is crucial for precise solutions.
• Practical applications include designing circuits and devices reliant on magnetic fields.