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Physics Part-I

Moving Charges and Magnetism

A toroid has a core (non -ferromagnetic) of inner radius 25 cm

Physics Part-I

Class 12

NCERT

Chapter 1: Electric Charges and Fields

47 questions

Chapter 2: Electrostatic Potential and Capacitance

46 questions

Chapter 3: Current Electricity

33 questions

Chapter 4: Moving Charges and Magnetism

41 questions

For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

B = \frac{μ _{0} I R ^{2} N}{2 ( x ^{2} + R ^{2} ) ^{3/2}}

(a) Show that this reduces to the familiar result for field at the centre of the coil.

(b) Consider two parallel co-axial circular coils of equal radius R,and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by.

B = 0.72 \frac{μ _{0} N I}{R}

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

A toroid has a core (non -ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.

Answer the following questions:

(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be setup to prevent the electron from deflecting from its straight line path.

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Solving time: 4 mins

A toroid has a core (non -ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.

Name: Video solution 1: [Solved] A toroid has a core (non -ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid. - Physics Part-I
Uploaded: 2022-06-13T12:33:48.000Z
Duration: 4 min
Description: Video solution 1: [Solved] A toroid has a core (non -ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid. - Physics Part-I

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Text solutionVerified

r_{1} = 0.25 m, r_{2} = 0.26 m, n = 3500, I = 11 A

(a)magnetic field outside is zero. It is non zero only inside the core of a toroid.

By definition, Ampere's Law states that integration of magnetic field over an amperian loop is equivalent to constant μ multiplied by current enclosed. Now, the current that flows at outer circumference and inner circumference of toroid are in opposite direction to each other. Therefore, current enclosed by Amperian loop is zero. Thus, magnetic field outside the toroid is zero.

(b)Magnetic field inside the core of a toroid is given by the relation

B = \frac{μ _{o} n I}{l}

l=length of toroid

where,

l = 2 \frac{π ( r _{1} + r _{2} )}{2} = .51 π

B = \frac{4 π \times 1 0 ^{- 7} \times 3500 \times 11}{0.51 π}

so,

B = 3 \times 1 0^{- 2} T

(c)Magnetic field in the empty space surrounded by the toroid is zero as there will no current in the empty space.

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Practice more questions from Physics Part-I (NCERT)

For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

B = \frac{μ _{0} I R ^{2} N}{2 ( x ^{2} + R ^{2} ) ^{3/2}}

(a) Show that this reduces to the familiar result for field at the centre of the coil.

B = 0.72 \frac{μ _{0} N I}{R}

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Answer the following questions:

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Question Text	A toroid has a core (non -ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.
Updated On	Jun 13, 2022
Topic	Moving Charges and Magnetism
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	214
Avg. Video Duration	4 min

Text solutionVerified

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