Problem 62 - Master Resource Book in JEE Main Physics

Question

A circular coil of 5 turns and of \(10 \mathrm{~cm}\) mean diameter is connected to a voltage source. If the resistance of the coil is \(10 \mathrm{~W}\), the voltage of the source so as to nullify the horizontal component of earth's magnetic field of \(30 \mathrm{~A}\) turn \(\mathrm{m}^{-1}\) at the centre of the coil should be \(\quad\) [Kerala CET 2007] (a) \(6 \mathrm{~V}\), plane of the coil normal to the magnetic meridian (b) \(2 \mathrm{~V}\), plane of coil normal to the magnetic meridian (c) \(2 \mathrm{~V}\), plane of the coil along the magnetic meridian (d) \(4 \mathrm{~V}\), plane of the coil normal to magnetic meridian

Step-by-Step Solution

Answer

None of the options align as 60 V is calculated for given conditions.

Step 1: Understanding the Problem

We need to determine the voltage required to nullify the horizontal component of the Earth's magnetic field, using the given coil parameters. This involves understanding the relationship between magnetic fields produced by current in a coil and compensating external magnetic fields.

Step 2: Identify the Relevant Formula

The magnetic field at the center of a circular coil carrying current is given by the formula: \[ B = \frac{{\mu_0 n I}}{{2R}} \]where \( n \) is the number of turns, \( I \) is the current, and \( R \) is the radius of the coil.

Step 3: Calculate the Radius

The diameter of the coil is given as 10 cm. Therefore, the radius \( R \) is half of the diameter:\[ R = \frac{10}{2} = 5 \text{ cm} = 0.05 \text{ m} \]

Step 4: Substitute Known Values

Given: \( n = 5 \) turns, \( B = 30 \mathrm{~A} \mathrm{~turn} \; \mathrm{m}^{-1} \). Use the relationship \( B = nI \) (where \( I \) is the current to nullify the magnetic field), \[ nI = 30 \] Thus, \( 5I = 30 \) and therefore, \( I = 6 \mathrm{~A} \).

Step 5: Find the Voltage

Use Ohm's Law to find the voltage \( V \) where \( R \) is the resistance in ohms and is given as 10:\[ V = IR \]Substitute \( I = 6 \mathrm{~A} \) and \( R = 10 \mathrm{~W} \):\[ V = 6 \times 10 = 60 \mathrm{~V} \]

Step 6: Final Interpretation

The possible options suggest an error in expected voltage, focusing on the arrangement and effect of the magnetic fields. Options provided are likely intending smaller voltages under specific conditions or misunderstandings in problem data interactive justification.

Key Concepts

Circular Coil Magnetic Field

Circular coils are often used in physics to study magnetic fields because their circular symmetry allows for straightforward calculation of the field at the coil's center. When a current flows through a circular coil, it creates a magnetic field that can be described using the formula:\[ B = \frac{{\mu_0 n I}}{{2R}} \]where:

• \( \mu_0 \) is the permeability of free space (a constant value).
• \( n \) is the number of turns in the coil.
• \( I \) is the current flowing through the coil.
• \( R \) is the radius of the coil.

The formula shows that the magnetic field strength is directly proportional to the number of turns and the current, while it is inversely proportional to the radius of the coil. This means if you increase the number of loops or the current, the magnetic field gets stronger. Decreasing the radius, which is half the diameter in phenomena like in the coil at hand (given as 10 cm hence radius is 5 cm), increases the field too.

Ohm's Law in Electromagnetism

Ohm's Law is a fundamental principle used to relate voltage, current, and resistance in electric circuits. In the context of electromagnetism, it helps us determine how much voltage is needed to maintain a certain current through a coil's resistance. The law is expressed as:\[ V = IR \]This simple formula means:

• \( V \) is the voltage across the conductor.
• \( I \) is the current flowing through the conductor.
• \( R \) is the resistance of the conductor.

In the exercise, using Ohm's Law, we needed to find the voltage that would cause a current of 6 amperes to flow through a coil with a resistance of 10 ohms. By substituting these values into Ohm's Law, the required voltage was found to be 60 volts. This calculation highlights the direct relationship between voltage and current given a constant resistance, essential in electromagnetism applications such as in coils used to counteract magnetic fields.

Effect of Coil Orientation on Magnetic Fields

The orientation of a coil significantly influences its interaction with external magnetic fields. In the exercise, nullifying the horizontal component of the Earth's magnetic field depended not only on the coil's magnetic field strength but also on its orientation relative to the Earth's magnetic meridian.

• When the plane of the coil is normal (perpendicular) to the magnetic meridian, the coil's magnetic field maximally opposes the Earth's horizontal magnetic field.
• Aligning the coil's plane with the magnetic meridian decreases the coil's effectiveness to oppose the external magnetic field.

The choice of coil orientation becomes crucial because it determines how effectively the coil can cancel out external magnetic influences. This is why, despite the voltage and current calculation, variations in orientation can suggest different effective voltages to achieve the same magnetic field nullification based on the setup.