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Physics Part-I

Moving Charges and Magnetism

For a circular coil of radius R and N turns carrying current I

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

A magnetic field of 100 G (1 G = 10

^{- 4}

T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about

1 0^{- 3} m^{2}

. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns

m^{- 1}

. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

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.

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Question

Medium

Solving time: 5 mins

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.]

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Benjamin Discussed

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.]

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

(a)

At the centre of the coil,

x = 0

Hence, magnetic field at centre is

B_{c} = \frac{μ _{o} I R ^{2} N}{2 R ^{3}}

B = \frac{μ _{o} N I}{2 R}

(b)

In a small region of length 2d about the mid-point between the coils,

B = \frac{μ _{0} I R ^{2} N}{2} \times ⎣ ⎡ {(\frac{R}{2} + d)^{2} + R^{2}}^{- 3/2} + {(\frac{R}{2} - d)^{2} + R^{2}}^{- 3/2} ⎦ ⎤

= \frac{μ _{0} I R ^{2} N}{2} \times (\frac{5 R ^{2}}{4})^{- 3/2} \times [(1 + \frac{4 d}{5 R}) + (1 - \frac{4 d}{5 R})^{- 3/2}]

= \frac{μ _{0} I R ^{2} N}{2} \times (\frac{4}{5})^{- 3/2} \times [1 - \frac{6 d}{5 R} + 1 + \frac{6 d}{5 R}]

where in the second and third steps above, terms containing

d^{2} / R^{2}

and higher powers of d/R are neglected since

\frac{d}{R} << 1

The terms linear in d/R cancel giving a uniform field B in a small region:

B = \frac{μ _{0} I R ^{2} N}{2} \times (\frac{4}{5})^{- 3/2} \frac{μ _{0} I N}{R} = 0.72 \frac{μ _{0} I N}{R}

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Practice more questions from Physics Part-I (NCERT)

A magnetic field of 100 G (1 G = 10

^{- 4}

T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about

1 0^{- 3} m^{2}

. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns

m^{- 1}

. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

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.]

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Question Text	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.]
Topic	Moving Charges and Magnetism
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	17

Text solutionVerified

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