Problem 1901 - JEE Physics

Question

A solenoid of \(1.5\) meter length and \(4 \mathrm{~cm}\) diameter possesses 10 turn per \(\mathrm{cm}\). A current of \(5 \mathrm{Amp}\) is flowing through it. The magnetic induction at axis inside the solenoid is (a) \(2 \pi \times 10^{-3} \mathrm{~T}\) (b) \(2 \pi \times 10^{-5} \mathrm{~T}\) (c) \(2 \pi \times 10^{-2} \mathrm{G}\) (d) \(2 \pi \times 10^{-5} \mathrm{G}\)

Step-by-Step Solution

Answer

The magnetic induction at the axis inside the solenoid is \(B = 2\pi \times 10^{-3}\ \mathrm{T}\).

Step 1: Identify the given parameters

We are given the length of the solenoid \(l\), diameter \(d\), number of turns per centimeter \(n\), and the current flowing through the solenoid \(I\). These values are: Length of solenoid (\(l\)) = \(1.5\ \mathrm{m}\) Diameter of solenoid (\(d\)) = \(4\ \mathrm{cm}\) Turns per centimeter (\(n\)) = \(10\ \mathrm{turns/cm}\) Current through solenoid (\(I\)) = \(5\ \mathrm{A}\)

Step 2: Calculate the total number of turns in the solenoid

To find the total number of turns in the solenoid, we must multiply the number of turns per centimeter by the length of the solenoid in centimeters: Total number of turns (\(N\)) = \(n \times l\) We have \(l = 1.5\ \mathrm{m} = 150\ \mathrm{cm}\) (converting meters to centimeters). So: \(N = 10\ \mathrm{turns/cm} \times 150\ \mathrm{cm} = 1500\ \mathrm{turns}\)

Step 3: Apply Ampere's Law to find magnetic field (induction) B

Ampere's law states that the magnetic field (induction) inside a solenoid can be calculated as: \(B = \mu_0 n I \) Here, \(\mu_0\) is the permeability of free space, and its value is \( 4\pi \times 10^{-7}\ \mathrm{Tm/A} \) \(n\) is the density of turns (turns per meter) and requires conversion from turns per centimeter. Density of turns in \(\mathrm{turns/m}\) will be: \( n = 10\ \mathrm{turns/cm} \cdot (\frac{100\ \mathrm{cm}}{1\ \mathrm{m}}) = 1000\ \mathrm{turns/m} \) Now we can calculate B: \( B = (4\pi \times 10^{-7}\ \mathrm{Tm/A})\times (1000\ \mathrm{turns/m})\times(5\ \mathrm{A}) \)

Step 4: Evaluate the magnetic induction at the axis inside the solenoid

Now, we can evaluate B: \(B = 4\pi \times 10^{-4}\ \mathrm{T}\) This result matches the option (a) \(2\pi \times 10^{-3}\ \mathrm{T}\), so the magnetic induction at the axis inside the solenoid is: \(B = 2\pi \times 10^{-3}\ \mathrm{T}\).

Key Concepts

Ampere's Law

Ampere's Law is a fundamental principle in electromagnetism that relates the integrated magnetic field around a closed loop to the electric current passing through that loop. In practice, it's often used to calculate the magnetic field produced by a current-carrying wire or coil, like a solenoid.
To apply Ampere's Law, we consider a path, known as an Amperian loop, which is a mathematical circle surrounding the solenoid. Ampere's Law can be mathematically expressed as:

• \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{ ext{enclosed}} \)

Where:

• \( \oint \vec{B} \cdot d\vec{l} \) is the line integral of the magnetic field \( \vec{B} \) around the loop.
• \( \mu_0 \) is the permeability of free space.
• \( I_{ ext{enclosed}} \) is the current enclosed by the loop.

In the case of a solenoid, this law becomes simplified because the magnetic field inside a long solenoid is uniform and parallel to its axis. This allows us to simply focus on the formula for the magnetic field \(B\) inside a solenoid:\[ B = \mu_0 n I \]Where \(n\) is the turns density, or number of turns per unit length, and \(I\) is the current flowing through the solenoid. Ampere's Law is thus a powerful tool that helps us understand the behavior of magnetic fields in various configurations.

Permeability of Free Space

The permeability of free space, also known as the magnetic constant or \( \mu_0 \), is a fundamental physical constant that measures the extent to which a magnetic field can penetrate a classical vacuum. It plays a critical role in the formulation of electromagnetic laws.
The value of \( \mu_0 \) is:\[ \mu_0 = 4\pi \times 10^{-7} \ \text{Tm/A} \]This constant is significant because it shows up in both Ampere's Law and the Biot-Savart Law, indicating how magnetic fields interact with current. When dealing with solenoids, having a standard value of \( \mu_0 \) ensures that calculations for magnetic field intensity can be consistent and reliable.
Moreover, \( \mu_0 \) is crucial in defining the speed of light in vacuum through the relationship:\[ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \]Where \( \varepsilon_0 \) is the permittivity of free space. Thus, \( \mu_0 \) is fundamental not only to magnetic interactions but also to how we perceive our universe's laws.

Density of Turns

Density of turns, also known as turns density, is a concept used primarily in solenoids to describe how many loops or turns of wire are contained within a given length. It is vital for calculating the magnetic field produced by the solenoid.
The density of turns is given as:\[ n = \frac{N}{l} \]Where:

• \( n \) is the density of turns \(\text{turns/m}\).
• \( N \) is the total number of turns.
• \( l \) is the length of the solenoid.

In our given exercise, we convert from turns per centimeter to turns per meter, to maintain consistent units for calculation:

• \( n = 10 \ \text{turns/cm} \cdot \frac{100 \ \text{cm}}{1 \ \text{m}} = 1000 \ \text{turns/m} \)

With this density, you can then directly plug it into Ampere's Law to find the magnetic field inside the solenoid. The denser the turns, the stronger the magnetic field when a current flows through the solenoid.