Q14 of 31 Page 170

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.

Here we have to find total magnetic field produced by the system so we will first find magnetic field due to each coil with direction and then add them in accordance with vector addition. Using the Right-hand thumb rule we can predict the direction of induced magnetic field in both the coils.


The orientation of both the coils is shown below in the figure.



The direction of induced magnetic field in both the coil can be found with the help of Right Hand Thumb Rule.


As Shown in figure.



So magnetic field in Coil X would be towards East, and in Coil Y would be towards West,


Explanation: if we curl our fingers in the Anticlockwise Sense in as in case of coil x our thumb will show the direction of the magnetic field which is towards East and if we curl our fingers in the clockwise Sense in as in case of coil Y our thumb will show the direction of the magnetic field which is towards West


The magnitude of magnetic field in a coil is given by the formula


Where, B is the magnetic field


I is the current in the coil


R is the radius of coil in metres


N is the number of turns


is the permittivity of free space = 4π × 10-7 TmA-1


For coil X


Current in the coil, Ix = 16A


Radius of the coil, Rx = 16cm = 0.16m


No. of turns of Coil, N �x = 20


So magnetic field due to coil X, Bx =


= 4π × 10-4 T


= 1.2510-3 T (Towards East)


For coil Y


Current in the coil, Iy = 18A


Radius of the coil, Ry = 10cm = 0.10m


No. of turns of Coil, N �y = 25


So magnetic field due to coil Y,


= 9π × 10-4 T


= 2.82 × 10-3 T (Towards West)


So, the Resultant magnetic field of the system = By- Bx


(since fields are in opposite directions)


= 2.82 × 10-3 T- 1.2510-3 T


= 1.57 × 10-3 T (Towards West)


So net magnetic field of the system is B = 1.57 × 10-3 T, Towards West direction


More from this chapter

All 31 →
13

A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

 13

Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

 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.

 16

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,


(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,


, approximately.


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