Problem 72 - Master Resource Book in JEE Main Physics

Question

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current \(i .\) The magnetic field at its centre is \(6.28 \times 10^{-2} \mathrm{Wbm}^{-2} .\) Another along solenoid has 100 turns per \(\mathrm{cm}\) and it carries a current \(i / 3 .\) The value of the magnetic field at its centre [AIEEE 2006] (a) \(1.05 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}\) (b) \(1.05 \times 10^{-2} \mathrm{~Wb} \mathrm{~m}^{-2}\) (c) \(1.05 \times 10^{-5} \mathrm{~Wb} \mathrm{~m}^{-2}\) (d) \(1.05 \times 10^{-3} \mathrm{~Wb} \mathrm{~m}^{-2}\)

Step-by-Step Solution

Answer

The magnetic field at the center of the second solenoid is \(1.05 \times 10^{-2} \mathrm{Wb/m^2}\), option (b).

Step 1: Understand the formula for the magnetic field in a solenoid

The magnetic field inside a long solenoid is given by the formula \( B = \mu_0 n i \), where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space (\(4\pi \times 10^{-7} \mathrm{Tm/A} \)), \( n \) is the number of turns per meter, and \( i \) is the current.

Step 2: Calculate the number of turns per meter for the second solenoid

The second solenoid has 100 turns per centimeter. Convert this to turns per meter: \( n = 100 \text{ turns/cm} = 100 \times 100 \text{ turns/m} = 10000 \text{ turns/m} \).

Step 3: Express the given data for the first solenoid

The first solenoid has a magnetic field \( B = 6.28 \times 10^{-2} \mathrm{Wb/m^2} \) with \( 200 \text{ turns/cm} = 200 \times 100 \text{ turns/m} = 20000 \text{ turns/m} \). The formula \( B = \mu_0 n i \) gives \( i = \frac{B}{\mu_0 n} = \frac{6.28 \times 10^{-2}}{4\pi \times 10^{-7} \times 20000} \).

Step 4: Calculate the current in the first solenoid

Substituting the values into the equation from Step 3, \( i = \frac{6.28 \times 10^{-2}}{4\pi \times 10^{-7} \times 20000} = \frac{6.28 \times 10^{-2}}{2.51 \times 10^{-2}} = 2.5 \times 10^{3} \mathrm{A} \).

Step 5: Calculate the current in the second solenoid

The second solenoid carries a current \( \frac{i}{3} = \frac{2500}{3} \).

Step 6: Calculate the magnetic field in the second solenoid

Using the formula \( B = \mu_0 n \frac{i}{3} \) for the second solenoid, \( B_2 = 4\pi \times 10^{-7} \times 10000 \times \frac{2500}{3} \).

Step 7: Simplify and solve for the magnetic field in the second solenoid

Calculating, \( B_2 = \frac{10^{-7} \times 10^4 \times 2500 \times 4\pi}{3} = \frac{10000 \mathrm{A/m} \times 2.5 \mathrm{A}}{3} = \frac{10^{4} \times 2.5 \times 4 \times 3.14159265}{3}\). Simplifying gives \( B_2 = 1.05 \times 10^{-2} \mathrm{Wb/m^2} \).

Step 8: Determine the correct answer

The magnetic field at the center of the second solenoid is \( 1.05 \times 10^{-2} \mathrm{Wb/m^2} \). Therefore, the correct answer is option (b).

Key Concepts

Permeability of Free Space

The permeability of free space, denoted by the symbol \( \mu_0 \), is a fundamental constant in physics. It represents how much resistance the vacuum of space offers against the formation of a magnetic field. The value is given as \( 4\pi \times 10^{-7} \ \text{Tm/A} \), where T stands for Tesla, m for meter, and A for Ampere.

This constant plays a crucial role in calculating the magnetic field within a solenoid. It acts as a scaling factor, directly influencing the strength of the magnetic field produced for a given current and number of turns per meter. Essentially, \( \mu_0 \) provides a baseline for understanding how efficiently a magnetic field can be established in free space. For instance, in the formula \( B = \mu_0 n i \), \( \mu_0 \) ensures that the relationship between the current, the number of turns, and the resulting magnetic field is consistent and predictable.

Understanding this constant is key to solving problems related to magnetic fields in solenoids, helping you determine how various factors such as changes in current or coil density affect the field's strength.

Turns per Meter

When discussing a solenoid, turns per meter \( n \) is a measure of how many coils of wire are wound around the solenoid per unit length. This is crucial because the density of these loops determines how strong the magnetic field inside the solenoid will be. Higher turns per meter mean more loops are packed into every meter of the solenoid, thus amplifying the magnetic field it produces.

In practical scenarios, this means a solenoid with a greater number of turns per meter will generate a stronger magnetic field than a solenoid with fewer turns, assuming all other factors such as current remain constant. For example, a solenoid with 200 turns/cm translates to \( 20000 \ \text{turns/m} \), yielding a formidable magnetic field compared to one with only \( 10000 \ \text{turns/m} \).

In formulas, turns per meter often appears as \( n \), and plays a pivotal role in the equation \( B = \mu_0 n i \), impacting the field's strength directly proportional to this density of loops. It's an essential factor for those seeking to optimize the efficacy of their electromagnets or any devices relying on solenoids.

Current in Solenoid

The current \( i \) flowing through a solenoid profoundly affects the strength of its magnetic field. More current means more charged particles move through the coils, increasing the interaction with the magnetic field lines and thereby strengthening the field.

In the equation \( B = \mu_0 n i \), the current directly propels the magnetic field's intensity. If you increase the current in a solenoid while keeping the turns per meter constant, the magnetic field's strength will rise proportionately. This relationship is crucial in applications like electromagnets or induction coils, where adjustable magnetic fields are required.

For example, in a solenoid that initially carries a current \( i \), reducing it to \( i/3 \) as shown in the exercise leads to a weaker magnetic field. The current must be tailored carefully depending on the desired outcome, whether the goal is to maximize the magnetic field or conserve power by running the solenoid at lower current levels. Understanding this concept is vital for fine-tuning the performance of devices that depend heavily on solenoidal magnetic fields.