Irodov Solutions: Constant Magnetic Field: Magnetics (3.5) | Physics Class 12 PDF Download

 Page 1


Constant Magnetic Field Magnetics (Part - 1) 
 
Q. 219. A current I = 1.00 A circulates in a round thin-wire loop of radius R = 100 
mm. Find the magnetic induction 
(a) at the centre of the loop;  
(b) at the point lying on the axis of the loop at a distance x = 100 mm from its 
centre. 
 
Solution. 219.  
 
 
 
 
 
 
 
(b) From Biota-Savart’s law :  
 
 
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Constant Magnetic Field Magnetics (Part - 1) 
 
Q. 219. A current I = 1.00 A circulates in a round thin-wire loop of radius R = 100 
mm. Find the magnetic induction 
(a) at the centre of the loop;  
(b) at the point lying on the axis of the loop at a distance x = 100 mm from its 
centre. 
 
Solution. 219.  
 
 
 
 
 
 
 
(b) From Biota-Savart’s law :  
 
 
 
 
 
And   
 
 
 
Here   is a unit vector perpendicular to the plane containing the current loop (Fig.) 
and in the direction of   
 
Thus we get   
 
Q. 220. A current I flows along a thin wire shaped as a regular polygon with n 
sides which can be inscribed into a circle of radius R. Find the magnetic induction 
at the centre of the polygon. Analyse the obtained expression n ? 8 
 
Solution. 220.  As   or perpendicular distance of any segment from  
 
centre equals  Now magnetic induction at O, due to die right current carrying 
element AB 
 
 
 
(From Biot- Savart’s law, the magnetic field at O due to any section such as AB is 
perpendicular to the plane of the figure and has the magnitude.) 
 
 
As there are n number of sides and magnetic induction vectors, due to each side at O, 
are equal in magnitude and direction. So, 
Page 3


Constant Magnetic Field Magnetics (Part - 1) 
 
Q. 219. A current I = 1.00 A circulates in a round thin-wire loop of radius R = 100 
mm. Find the magnetic induction 
(a) at the centre of the loop;  
(b) at the point lying on the axis of the loop at a distance x = 100 mm from its 
centre. 
 
Solution. 219.  
 
 
 
 
 
 
 
(b) From Biota-Savart’s law :  
 
 
 
 
 
And   
 
 
 
Here   is a unit vector perpendicular to the plane containing the current loop (Fig.) 
and in the direction of   
 
Thus we get   
 
Q. 220. A current I flows along a thin wire shaped as a regular polygon with n 
sides which can be inscribed into a circle of radius R. Find the magnetic induction 
at the centre of the polygon. Analyse the obtained expression n ? 8 
 
Solution. 220.  As   or perpendicular distance of any segment from  
 
centre equals  Now magnetic induction at O, due to die right current carrying 
element AB 
 
 
 
(From Biot- Savart’s law, the magnetic field at O due to any section such as AB is 
perpendicular to the plane of the figure and has the magnitude.) 
 
 
As there are n number of sides and magnetic induction vectors, due to each side at O, 
are equal in magnitude and direction. So, 
  
 
 
 
Q. 221. Find the magnetic induction at the centre of a rectangular wire frame 
whose diagonal is equal to d = 16 cm and the angle between the diagonals is equal 
to f = 30°; the current flowing in the frame equals I = 5.0 A. 
 
Solution. 221. We know that magnetic induction due to a straight current carrying wire 
at any point, at a perpendicular distance from it is given by : 
 
 
 
where r is the perpendicular distance of the wire from the point, considered, and ? 1 is 
the angle between the line, joining the upper point of straight wire to the considered 
point and the perpendicular drawn to the wire and ? 2 that from the lower point of the 
straight wire. 
 
Here,   
 
And    
Page 4


Constant Magnetic Field Magnetics (Part - 1) 
 
Q. 219. A current I = 1.00 A circulates in a round thin-wire loop of radius R = 100 
mm. Find the magnetic induction 
(a) at the centre of the loop;  
(b) at the point lying on the axis of the loop at a distance x = 100 mm from its 
centre. 
 
Solution. 219.  
 
 
 
 
 
 
 
(b) From Biota-Savart’s law :  
 
 
 
 
 
And   
 
 
 
Here   is a unit vector perpendicular to the plane containing the current loop (Fig.) 
and in the direction of   
 
Thus we get   
 
Q. 220. A current I flows along a thin wire shaped as a regular polygon with n 
sides which can be inscribed into a circle of radius R. Find the magnetic induction 
at the centre of the polygon. Analyse the obtained expression n ? 8 
 
Solution. 220.  As   or perpendicular distance of any segment from  
 
centre equals  Now magnetic induction at O, due to die right current carrying 
element AB 
 
 
 
(From Biot- Savart’s law, the magnetic field at O due to any section such as AB is 
perpendicular to the plane of the figure and has the magnitude.) 
 
 
As there are n number of sides and magnetic induction vectors, due to each side at O, 
are equal in magnitude and direction. So, 
  
 
 
 
Q. 221. Find the magnetic induction at the centre of a rectangular wire frame 
whose diagonal is equal to d = 16 cm and the angle between the diagonals is equal 
to f = 30°; the current flowing in the frame equals I = 5.0 A. 
 
Solution. 221. We know that magnetic induction due to a straight current carrying wire 
at any point, at a perpendicular distance from it is given by : 
 
 
 
where r is the perpendicular distance of the wire from the point, considered, and ? 1 is 
the angle between the line, joining the upper point of straight wire to the considered 
point and the perpendicular drawn to the wire and ? 2 that from the lower point of the 
straight wire. 
 
Here,   
 
And    
 
 
Hence, the magnitude of total magnetic induction at O, 
 
 
 
Q. 222. A current l = 5.0 A flows along a thin wire shaped as shown in Fig. 3.59. 
The radius of a curved part of the wire is equal to R = 120 mm, the angle 2f = 90°. 
Find the magnetic induction of the field at the point O.  
 
 
 
Solution. 222. Magnetic induction due to the arc segment at O, 
 
 
 
 
And magnetic induction due to the line segment at O, 
Page 5


Constant Magnetic Field Magnetics (Part - 1) 
 
Q. 219. A current I = 1.00 A circulates in a round thin-wire loop of radius R = 100 
mm. Find the magnetic induction 
(a) at the centre of the loop;  
(b) at the point lying on the axis of the loop at a distance x = 100 mm from its 
centre. 
 
Solution. 219.  
 
 
 
 
 
 
 
(b) From Biota-Savart’s law :  
 
 
 
 
 
And   
 
 
 
Here   is a unit vector perpendicular to the plane containing the current loop (Fig.) 
and in the direction of   
 
Thus we get   
 
Q. 220. A current I flows along a thin wire shaped as a regular polygon with n 
sides which can be inscribed into a circle of radius R. Find the magnetic induction 
at the centre of the polygon. Analyse the obtained expression n ? 8 
 
Solution. 220.  As   or perpendicular distance of any segment from  
 
centre equals  Now magnetic induction at O, due to die right current carrying 
element AB 
 
 
 
(From Biot- Savart’s law, the magnetic field at O due to any section such as AB is 
perpendicular to the plane of the figure and has the magnitude.) 
 
 
As there are n number of sides and magnetic induction vectors, due to each side at O, 
are equal in magnitude and direction. So, 
  
 
 
 
Q. 221. Find the magnetic induction at the centre of a rectangular wire frame 
whose diagonal is equal to d = 16 cm and the angle between the diagonals is equal 
to f = 30°; the current flowing in the frame equals I = 5.0 A. 
 
Solution. 221. We know that magnetic induction due to a straight current carrying wire 
at any point, at a perpendicular distance from it is given by : 
 
 
 
where r is the perpendicular distance of the wire from the point, considered, and ? 1 is 
the angle between the line, joining the upper point of straight wire to the considered 
point and the perpendicular drawn to the wire and ? 2 that from the lower point of the 
straight wire. 
 
Here,   
 
And    
 
 
Hence, the magnitude of total magnetic induction at O, 
 
 
 
Q. 222. A current l = 5.0 A flows along a thin wire shaped as shown in Fig. 3.59. 
The radius of a curved part of the wire is equal to R = 120 mm, the angle 2f = 90°. 
Find the magnetic induction of the field at the point O.  
 
 
 
Solution. 222. Magnetic induction due to the arc segment at O, 
 
 
 
 
And magnetic induction due to the line segment at O, 
 
 
So, total magnetic induction at O, 
 
 
 
Q. 223. Find the magnetic induction of the field at the point O of a loop with 
current I, whose shape is illustrated 
(a) in Fig. 3.60a, the radii a and b, as well as the angle f are known; 
(b) in Fig. 3.60b, the radius a and the side b are known.  
 
 
 
Solution. 223. (a) From the Biot-Savart law, 
 
 
 
So, magnetic field induction due to the segment 1 at O, 
 
 
 
Also   
 
And