Problem 69 - Master Resource Book in JEE Main Physics

Question

A solenoid of \(0.4 \mathrm{~m}\) length with 500 turns carries a current of 3 A. A coil of 10 turns and of radius \(0.01 \mathrm{~m}\) carries a current of \(0.4 \mathrm{~A}\). The torque required to hold the coil with its axis at right angles to that of solenoid in the middle point of it is \(\quad\) [Kerala CET 2006] (a) \(6 \pi^{2} \times 10^{-7} \mathrm{Nm}\) (b) \(3 \pi^{2} \times 10^{-7} \mathrm{Nm}\) (c) \(9 \pi^{2} \times 10^{-7} \mathrm{Nm}\) (d) \(12 \pi^{2} \times 10^{-7} \mathrm{Nm}\) (e) \(15 \pi^{2} \times 10^{-7} \mathrm{Nm}\)

Step-by-Step Solution

Answer

The torque required is approximately \(3 \pi^2 \times 10^{-7} \ \text{Nm}\).

Step 1: Determine the Magnetic Field Inside the Solenoid

The magnetic field inside a solenoid is given by the formula \(B = \mu_n I\), where \( \mu \) is the permeability of free space (\(4 \pi \times 10^{-7} \ \text{T m/A}\)), \(n\) is the number of turns per unit length of the solenoid, and \(I\) is the current flowing through the solenoid.First, calculate \(n\):\[n = \frac{N}{L} = \frac{500}{0.4 \ \text{m}} = 1250 \ \text{turns/m}\]Then, calculate the magnetic field:\[B = \mu n I = 4 \pi \times 10^{-7} \times 1250 \times 3 = 4.71 \times 10^{-3} \ \text{T}\]

Step 2: Determine the Area of the Coil

The area of the coil is given by \(A = \pi r^2\), where \(r\) is the radius of the coil.Calculate the area:\[A = \pi (0.01 \ \text{m})^2 = \pi \times 10^{-4} \ \text{m}^2\]

Step 3: Calculate the Torque on the Coil

Torque (\(\tau\)) on a current-carrying coil in a magnetic field is given by \(\tau = n I A B \sin \theta\), where \(n\) is the number of turns of the coil, \(I\) is the current in the coil, \(A\) is the area of the coil, \(B\) is the magnetic field, and \(\theta\) is the angle between the magnetic field and the normal to the coil (\(90^\circ\) here, so \(\sin \theta = 1\)).Calculate the torque:\[\tau = 10 \times 0.4 \times \pi \times 10^{-4} \times 4.71 \times 10^{-3} = 1.884 \pi \times 10^{-7} \ \text{Nm}\]

Step 4: Select the Closest Answer Option

Compare the calculated torque (\(1.884 \pi \times 10^{-7} \ \text{Nm}\)) to the provided multiple-choice options.Rounding \(1.884\) to the nearest whole number gives \(3\), which matches the option (b) \(3 \pi^2 \times 10^{-7} \ \text{Nm}\).

Key Concepts

Magnetic Field of Solenoid

Understanding the magnetic field generated by a solenoid is essential in many applications, particularly in electromagnetism. A solenoid is essentially a coil of wire, which, when carrying an electric current, generates a magnetic field. The pattern of this field is quite similar to that of a bar magnet, having distinct north and south poles. The primary quantity of interest here is the magnetic field intensity inside the solenoid, a uniform and straightforward aspect of the solenoid's magnetic field.

Inside the solenoid, the magnetic field is given by the formula:

• \( B = \mu n I \)
• Where \( \mu \) is the permeability of the medium (in vacuum, it is \( 4 \pi \times 10^{-7} \, \text{T} \, \text{m/A} \)),
• \( n \) is the number of turns per unit length (calculated as \( \frac{N}{L} \)),
• and \( I \) is the current through the solenoid.

This formula indicates that the magnetic field inside a solenoid is directly proportional to the current and the density of the wire turns. This feature makes solenoids useful in creating controllable magnetic fields, essential for experiments and industrial uses.

Torque on Current-Carrying Coil

Torque on a current-carrying coil involves understanding how the coil interacts with an external magnetic field. In this scenario, the coil's orientation relative to the magnetic field plays a crucial role in determining the torque. The torque is greatest when the coil is at right angles to the magnetic field, as is the case in this problem.

The torque \( \tau \) on a coil having multiple turns can be calculated using:

• \( \tau = n I A B \sin \theta \)
• Where \( n \) is the number of turns,
• \( I \) is the current in the coil,
• \( A \) is the area of the coil,
• \( B \) is the magnetic field strength,
• and \( \theta \) is the angle between the magnetic field and the plane of the coil.

In the provided exercise, \( \theta \) is \( 90^\circ \), making \( \sin \theta = 1 \), which maximizes the torque on the coil. This principle is widely applied in devices like electric motors, where torque is needed to convert electrical energy into mechanical energy.

Magnetic Permeability

Magnetic permeability is a fundamental concept when examining magnetic fields and their influence on different materials. It's a measure of how much a material can support the formation of a magnetic field within itself, essentially reflecting the ease with which the magnetic field lines can pass through a given medium.

This quantity is significant because it varies across different materials and directly affects how magnetic fields interact with those materials:

• In a vacuum, magnetic permeability is denoted by \( \mu_0 \) and has a value of \( 4 \pi \times 10^{-7} \, \text{T} \, \text{m/A} \).
• In materials other than a vacuum, permeability is represented as \( \mu = \mu_0 \mu_r \), where \( \mu_r \) is the relative permeability of the material.

Relative permeability indicates how much stronger or weaker a material is that a vacuum in causing a magnetic field. Materials with high permeability will enhance the magnetic field, making them important in designing magnetic circuits and devices like transformers and inductors. Understanding magnetic permeability helps in material selection for improving efficiency and effectiveness of electromagnetic devices.