Home
Class 12
PHYSICS
A wire of length l is used to form a coi...

A wire of length l is used to form a coil. The magnetic field at its centre for a given current in it is minimum if the coil has

A

`4 turn`

B

`2 turn`

C

`1 turn`

D

data is not sufficient

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the number of turns in a coil that minimizes the magnetic field at its center for a given current, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Problem**: We have a wire of length \( L \) that is formed into a coil, and we need to determine how many turns (n) will minimize the magnetic field at the center of the coil for a given current \( I \). 2. **Relate Length of Wire to Turns**: When the wire is formed into a coil with \( n \) turns, the total length of the wire is equal to the circumference of the coil multiplied by the number of turns. The circumference of a circular loop is given by \( 2\pi r \), where \( r \) is the radius of the coil. Therefore, we can write: \[ L = n \cdot 2\pi r \] From this, we can express the radius \( r \) in terms of \( L \) and \( n \): \[ r = \frac{L}{2\pi n} \] 3. **Magnetic Field at the Center of the Coil**: The magnetic field \( B \) at the center of a circular coil with \( n \) turns carrying current \( I \) is given by the formula: \[ B = \frac{\mu_0 n I}{2r} \] Substituting the expression for \( r \) from the previous step, we get: \[ B = \frac{\mu_0 n I}{2 \left( \frac{L}{2\pi n} \right)} = \frac{\mu_0 n I \cdot 2\pi n}{2L} = \frac{\mu_0 \pi n^2 I}{L} \] 4. **Analyze the Magnetic Field**: From the equation \( B = \frac{\mu_0 \pi n^2 I}{L} \), we can see that the magnetic field \( B \) is directly proportional to \( n^2 \). This means that as the number of turns \( n \) increases, the magnetic field at the center also increases. 5. **Minimize the Magnetic Field**: To minimize the magnetic field \( B \), we need to minimize \( n \). The minimum value of \( n \) in the given options is \( 1 \) turn. 6. **Conclusion**: Therefore, the magnetic field at the center of the coil is minimum when the coil has **1 turn**. ### Final Answer: The coil should have **1 turn** to minimize the magnetic field at its center for a given current. ---

To solve the problem of finding the number of turns in a coil that minimizes the magnetic field at its center for a given current, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Problem**: We have a wire of length \( L \) that is formed into a coil, and we need to determine how many turns (n) will minimize the magnetic field at the center of the coil for a given current \( I \). 2. **Relate Length of Wire to Turns**: When the wire is formed into a coil with \( n \) turns, the total length of the wire is equal to the circumference of the coil multiplied by the number of turns. The circumference of a circular loop is given by \( 2\pi r \), where \( r \) is the radius of the coil. Therefore, we can write: \[ ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SOURCES OF MAGNETIC FIELD

    CENGAGE PHYSICS ENGLISH|Exercise Exercise (multiple Currect )|5 Videos
  • SOURCES OF MAGNETIC FIELD

    CENGAGE PHYSICS ENGLISH|Exercise Exercise (assertion-reasioning )|2 Videos
  • SOURCES OF MAGNETIC FIELD

    CENGAGE PHYSICS ENGLISH|Exercise Exercise (subjective )|10 Videos
  • RAY OPTICS

    CENGAGE PHYSICS ENGLISH|Exercise DPP 1.6|12 Videos
  • WAVE OPTICS

    CENGAGE PHYSICS ENGLISH|Exercise Comprehension Type|14 Videos

Similar Questions

Explore conceptually related problems

The magnetic field at the centre of the current carrying coil is

A wire of certain length carries a current l.It is bent to form a circle of one turn and the magnetic field at the centre is B1. If it is bent to form a coil of four turns, then magnetic field at centre is B_2 .The ratio of B_1 and B_2 is

Two coils are having magnetic field B and 2B at their centres and current i and 2i then the ratio of their radius is

The magnetic field at centre of a hexagonal coil of side l carrying a current I is

A circular coil of radius R carries a current i . The magnetic field at its centre is B . The distance from the centre on the axis of the coil where the magnetic field will be B//8 is

Ratio of magnetic field at the centre of a current carrying coil of radius R and at a distance of 3R on its axis is

A conductor of length L and carrying current i is bent to form a coil of two turns.Magnetic field at the centre of the coil will be

A circular coil of 100 turns has a radius of 10 cm and carries a current of 5 A. Determine the magnetic field at the centre of the coil and at a point on the axis of the coil at a distance of 5 cm from its centre.

A wire of length 2 m carrying a current of 1 A is bend to form a circle. The magnetic moment of the coil is (in Am^(2) )

The ratio of the magnetic field at the centre of a current carrying circular coil to its magnetic moment is x. If the current and radius both are doubled the new ratio will become