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Page No 170: - Chapter 4 Moving Charges & Magnetism Additional Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]


Updated By SaraNextGen
On April 24, 2024, 11:35 AM

Question 4.14:

Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Answer:

Radius of coil X, r1 = 16 cm = 0.16 m

Radius of coil Y, r2 = 10 cm = 0.1 m

Number of turns of on coil X, n1 = 20

Number of turns of on coil Y, n2 = 25

Current in coil X, I1 = 16 A

Current in coil Y, I2 = 18 A

Magnetic field due to coil X at their centre is given by the relation,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6934/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m5db54452.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6934/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m144f80ba.gif = Permeability of free space = https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6934/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m7ae7f5fe.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6934/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_47fedc59.gif

Magnetic field due to coil Y at their centre is given by the relation,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6934/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_228752b7.gif

Hence, net magnetic field can be obtained as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6934/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m4b34c84a.gif

Question 4.15:

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 10−3 m2. 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

Answer:

Magnetic field strength, B = 100 G = 100 × 10−4 T

Number of turns per unit length, n = 1000 turns m−1

Current flowing in the coil, I = 15 A

Permeability of free space, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6936/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m144f80ba.gifhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6936/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m7ae7f5fe.gif

Magnetic field is given by the relation,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6936/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_17cc1ebd.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6936/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_md25a31e.gif

If the length of the coil is taken as 50 cm, radius 4 cm, number of turns 400, and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.

Question 4.16:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m102d2e23.gif

(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,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m2f06bad5.gif, approximately.

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

Answer:

Radius of circular coil = R

Number of turns on the coil = N

Current in the coil = I

Magnetic field at a point on its axis at distance x is given by the relation,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m102d2e23.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m144f80ba.gif = Permeability of free space

(a) If the magnetic field at the centre of the coil is considered, then x = 0.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m6e0742da.gif

This is the familiar result for magnetic field at the centre of the coil.

(b) Radii of two parallel co-axial circular coils = R

Number of turns on each coil = N

Current in both coils = I

Distance between both the coils = R

Let us consider point Q at distance d from the centre.

Then, one coil is at a distance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m4e2bb696.giffrom point Q.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_4dd19828.gifMagnetic field at point Q is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m60e98ed7.gif

Also, the other coil is at a distance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m527af0c3.giffrom point Q.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_4dd19828.gifMagnetic field due to this coil is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_627bb1e.gif

Total magnetic field,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m791fb594.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6938/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m2c6ddc52.gif

Hence, it is proved that the field on the axis around the mid-point between the coils is uniform.

Question 4.17:

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:

Inner radius of the toroid, r1 = 25 cm = 0.25 m

Outer radius of the toroid, r2 = 26 cm = 0.26 m

Number of turns on the coil, N = 3500

Current in the coil, I = 11 A

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

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

B = https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6941/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_5ccae21a.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6941/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m144f80ba.gif = Permeability of free space = https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6941/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_m7ae7f5fe.gif

l = length of toroid

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/248/6941/NS_17-11-2008_Sravana_12_Physics_4_28_NRJ_SG_html_5208eed0.gif

(c) Magnetic field in the empty space surrounded by the toroid is zero.

Question 4.18:

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 set up to prevent the electron from deflecting from its straight line path.

Answer:

(a) The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field.

(b) Yes, the final speed of the charged particle will be equal to its initial speed. This is because magnetic force can change the direction of velocity, but not its magnitude.

(c) An electron travelling from West to East enters a chamber having a uniform electrostatic field in the North-South direction. This moving electron can remain undeflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the South. According to Fleming’s left hand rule, magnetic field should be applied in a vertically downward direction.

Also Read : Page-No-171:-Chapter-4-Moving-Charges-&-Magnetism-Additional-Exercise-Solutions-class-12-ncert-solutions-Physics

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