Problem 1886 - JEE Physics

Question

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current of \(2.5 \mathrm{Amp}\). The mag. field at its centre is tesla. (a) \(\pi \times 10^{-2}\) (b) \(2 \pi \times 10^{-2}\) (c) \(3 \pi \times 10^{-2}\) (d) \(4 \pi \times 10^{-2}\)

Step-by-Step Solution

Answer

The magnetic field at the center of the solenoid is (a) \(\pi \times 10^{-2} \mathrm{T}\).

Step 1: Convert turns per centimeter to turns per meter

\ First, we will need to convert the given number of turns per centimeter to turns per meter. Recall that there are 100 centimeters in 1 meter. So, \(n = 200 \text{ turns/cm} \times 100 \text{ cm/m} = 20000 \text{ turns/m}\).

Step 2: Recall the formula for the magnetic field inside a long solenoid

\ The formula for the magnetic field inside a long solenoid is \(B = \mu_0 n I\), where \(B\) is the magnetic field, \(\mu_0\) is the vacuum permeability constant (\(4\pi \times 10^{-7} \text{ T m/A}\)), \(n\) is the number of turns per meter, and \(I\) is the current.

Step 3: Plug in the values and solve for the magnetic field

\ Now, plug the given values into the formula and solve for the magnetic field: \(B = (4\pi \times 10^{-7} \text{ T m/A}) \times (20000 \text{ turns/m}) \times (2.5 \mathrm{A})\).

Step 4: Calculate the magnetic field

\ Using a calculator, compute the magnetic field: \(B = (4\pi \times 10^{-7}) \times (20000) \times (2.5)\) \(B = \pi \times 10^{-2} \mathrm{T}\).

Step 5: Compare the result to the given options

\ Our calculated magnetic field is \(\pi \times 10^{-2} \text{ T}\), which matches option (a). Therefore, the correct answer is (a) \(\pi \times 10^{-2} \text{ T}\).

Key Concepts

Turns per Meter

When dealing with solenoids, understanding the concept of "turns per meter" is crucial. A solenoid is a coil of wire, and the "turns" refer to the loops of the wire around the solenoid. 'Turns per meter' is a way to express how tightly these loops are wound.
To calculate the turns per meter from turns per centimeter, multiply the number of turns by the number of centimeters in a meter, which is 100. In our example, 200 turns/cm converted to turns per meter becomes:

• 200 turns/cm \( \times 100 \, \text{cm/m} = 20000 \, \text{turns/m} \)

Understanding this conversion allows us to apply calculations directly to the solenoid's physical dimensions. The tighter the winding, the stronger the magnetic field generated in the solenoid.

Vacuum Permeability Constant

The vacuum permeability constant, denoted as \( \mu_0 \), is fundamental when dealing with magnetic fields. It's a physical constant that represents the ability of a vacuum to support magnetic fields. The value of this constant is \( 4\pi \times 10^{-7} \, \text{T m/A} \).
This constant appears in many electromagnetism equations, helping quantify how much a material can pass through a magnetic field. For solenoids, \( \mu_0 \) is used in the magnetic field formula to determine the field's intensity.
Keep in mind, materials other than vacuum modify this permeability value, but for basic calculations and understanding, \( \mu_0 \) provides a simplified model.

Current in a Solenoid

The current (I) flowing through a solenoid is a vital factor influencing the magnetic field's strength inside it. Current, measured in amperes (A), causes a magnetic field to arise within the solenoid's core.
The more current flowing through the solenoid's wire, the stronger the magnetic field generated. Thus, in our example:

• Current \( I = 2.5 \, \text{A} \)

This current, combined with the turns per meter and the vacuum permeability constant, determines the net magnetic field inside the solenoid. The relation between these values is linear; doubling the current doubles the magnetic field strength, making current control important in magnetic applications.

Magnetic Field Formula

The magnetic field inside a long solenoid can be calculated using a straightforward formula: \( B = \mu_0 n I \). Here:

• \( B \) represents the magnetic field, measured in teslas (T).
• \( \mu_0 \) is the vacuum permeability constant.
• \( n \) is the number of turns per meter.
• \( I \) is the current in amperes.

This formula allows us to calculate the field's strength based on physical and electrical characteristics of the solenoid. Plugging in the numbers from our example:

• \( B = (4\pi \times 10^{-7} \, \text{T m/A}) \times (20000 \, \text{turns/m}) \times (2.5 \, \text{A}) = \pi \times 10^{-2} \, \text{T} \)

With this formula, you can determine how different factors affect the magnetic field within a solenoid, proving invaluable in both academic and practical applications.