Solved Example: Magnetic Effect of Electric Current and Magentism | Physics for JEE Main & Advanced PDF Download

 Page 1


Solved Example on Magnetic Effect of Electric Current 
and Magnetism 
JEE Mains 
Q1: Magnetic field due to a current carrying loop or a coil at a distant axial point ?? is ?? ?? and at an 
equal distance in it's plane is ?? ?? then 
?? ?? ?? ?? is 
(a) 2 
(b) 1 
(c) 
?? ?? 
(d) None of these 
Ans:  Current carrying coil behaves as a bar magnet as shown in figure. 
 
We also knows for a bar magnet, if axial and equatorial distance are same then ?? ?? = 2?? ?? 
Hence, in this equation 
?? 1
?? 2
=
2
1
 
Q2: Find the position of point from wire ' ?? ' where net magnetic field is zero due to following 
current distribution 
(a) ?? ???? 
(b) 
????
?? ????                                                                                           
(c) 
????
?? ???? 
(d) ?? ???? 
Page 2


Solved Example on Magnetic Effect of Electric Current 
and Magnetism 
JEE Mains 
Q1: Magnetic field due to a current carrying loop or a coil at a distant axial point ?? is ?? ?? and at an 
equal distance in it's plane is ?? ?? then 
?? ?? ?? ?? is 
(a) 2 
(b) 1 
(c) 
?? ?? 
(d) None of these 
Ans:  Current carrying coil behaves as a bar magnet as shown in figure. 
 
We also knows for a bar magnet, if axial and equatorial distance are same then ?? ?? = 2?? ?? 
Hence, in this equation 
?? 1
?? 2
=
2
1
 
Q2: Find the position of point from wire ' ?? ' where net magnetic field is zero due to following 
current distribution 
(a) ?? ???? 
(b) 
????
?? ????                                                                                           
(c) 
????
?? ???? 
(d) ?? ???? 
Ans: Suppose ?? is the point between the conductors where net magnetic field is zero. 
So at ?? | Magnetic field due to conductor 1| = | Magnetic field due to conductor 2 | 
i.e. 
?? 0
4?? ·
2(5?? )
?? =
?? 0
4?? ·
2(2?? )
(6-?? )
?
5
?? =
9
6-?? ? ?? =
30
7
 cm  
Hence position from ?? = 6 -
30
7
=
12
7
 cm 
Q3: A wire of fixed length is turned to form a coil of one turn. It is again turned to form a coil of 
three turns. If in both cases same amount of current is passed, then the ratio of the intensities of 
magnetic field produced at the centre of a coil will be 
(a) 9 times of first case 
(b) 
?? ?? times of first case (c) 3 times of first case 
(d) 
?? ?? times of first case 
Ans: Magnetic field at the centre of ?? turn coil carrying current ?? 
?? =
?? 0
4?? ·
2?????? ?? 
For single turn ?? = 1 
?? =
?? 0
4?? ·
2????
?? 
If the same wire is turn again to form a coil of three turns i.e. ?? = 3 and radius of each turn ?? '
=
?? 3
 
So new magnetic field at centre ?? '
=
?? 0
4?? ·
2?? (3)
?? '
 ? ?? '
= 9 ×
?? 0
4?? ·
2????
?? 
Comparing equation (ii) and (iii) gives ?? '
= 9?? . 
Q4:  An equilateral triangle of side 'a' carries a current ?? then find out the magnetic field at point ?? 
which is vertex of triangle 
(a) 
?? ?? ?? ?? v?? ????
? 
(b) 
?? ?? ?? ?? v?? ????
? 
(c) 
?? v?? ?? ?? ?? ????
?                                                                                                      
Page 3


Solved Example on Magnetic Effect of Electric Current 
and Magnetism 
JEE Mains 
Q1: Magnetic field due to a current carrying loop or a coil at a distant axial point ?? is ?? ?? and at an 
equal distance in it's plane is ?? ?? then 
?? ?? ?? ?? is 
(a) 2 
(b) 1 
(c) 
?? ?? 
(d) None of these 
Ans:  Current carrying coil behaves as a bar magnet as shown in figure. 
 
We also knows for a bar magnet, if axial and equatorial distance are same then ?? ?? = 2?? ?? 
Hence, in this equation 
?? 1
?? 2
=
2
1
 
Q2: Find the position of point from wire ' ?? ' where net magnetic field is zero due to following 
current distribution 
(a) ?? ???? 
(b) 
????
?? ????                                                                                           
(c) 
????
?? ???? 
(d) ?? ???? 
Ans: Suppose ?? is the point between the conductors where net magnetic field is zero. 
So at ?? | Magnetic field due to conductor 1| = | Magnetic field due to conductor 2 | 
i.e. 
?? 0
4?? ·
2(5?? )
?? =
?? 0
4?? ·
2(2?? )
(6-?? )
?
5
?? =
9
6-?? ? ?? =
30
7
 cm  
Hence position from ?? = 6 -
30
7
=
12
7
 cm 
Q3: A wire of fixed length is turned to form a coil of one turn. It is again turned to form a coil of 
three turns. If in both cases same amount of current is passed, then the ratio of the intensities of 
magnetic field produced at the centre of a coil will be 
(a) 9 times of first case 
(b) 
?? ?? times of first case (c) 3 times of first case 
(d) 
?? ?? times of first case 
Ans: Magnetic field at the centre of ?? turn coil carrying current ?? 
?? =
?? 0
4?? ·
2?????? ?? 
For single turn ?? = 1 
?? =
?? 0
4?? ·
2????
?? 
If the same wire is turn again to form a coil of three turns i.e. ?? = 3 and radius of each turn ?? '
=
?? 3
 
So new magnetic field at centre ?? '
=
?? 0
4?? ·
2?? (3)
?? '
 ? ?? '
= 9 ×
?? 0
4?? ·
2????
?? 
Comparing equation (ii) and (iii) gives ?? '
= 9?? . 
Q4:  An equilateral triangle of side 'a' carries a current ?? then find out the magnetic field at point ?? 
which is vertex of triangle 
(a) 
?? ?? ?? ?? v?? ????
? 
(b) 
?? ?? ?? ?? v?? ????
? 
(c) 
?? v?? ?? ?? ?? ????
?                                                                                                      
 
(d) Zero 
Ans:  
Solution: (b) As shown in the following figure magnetic field at ?? due to side 1 and side 2 is zero. 
 
Magnetic field at ?? is only due to side 3, which is ?? 1
=
?? 0
4?? ·
2?? sin 30
°
v3?? 2
? 
=
?? 0
4?? ·
2?? v3?? ?=
?? 0
?? 2v3????
? 
 
Q5:  A battery is connected between two points ?? and ?? on the circumference of a uniform 
conducting ring of radius ?? and resistance ?? . One of the arcs ???? of the ring subtends an angle ?? at 
the centre. The value of, the magnetic induction at the centre due to the current in the ring is 
(a) 
Proportional to ?? (??????
°
- ?? ) 
(b) 
Inversely proportional 
to ?? 
(c) Zero, only if ?? = ??????
°
 
(d) 
Zero for all values of 
?? 
Page 4


Solved Example on Magnetic Effect of Electric Current 
and Magnetism 
JEE Mains 
Q1: Magnetic field due to a current carrying loop or a coil at a distant axial point ?? is ?? ?? and at an 
equal distance in it's plane is ?? ?? then 
?? ?? ?? ?? is 
(a) 2 
(b) 1 
(c) 
?? ?? 
(d) None of these 
Ans:  Current carrying coil behaves as a bar magnet as shown in figure. 
 
We also knows for a bar magnet, if axial and equatorial distance are same then ?? ?? = 2?? ?? 
Hence, in this equation 
?? 1
?? 2
=
2
1
 
Q2: Find the position of point from wire ' ?? ' where net magnetic field is zero due to following 
current distribution 
(a) ?? ???? 
(b) 
????
?? ????                                                                                           
(c) 
????
?? ???? 
(d) ?? ???? 
Ans: Suppose ?? is the point between the conductors where net magnetic field is zero. 
So at ?? | Magnetic field due to conductor 1| = | Magnetic field due to conductor 2 | 
i.e. 
?? 0
4?? ·
2(5?? )
?? =
?? 0
4?? ·
2(2?? )
(6-?? )
?
5
?? =
9
6-?? ? ?? =
30
7
 cm  
Hence position from ?? = 6 -
30
7
=
12
7
 cm 
Q3: A wire of fixed length is turned to form a coil of one turn. It is again turned to form a coil of 
three turns. If in both cases same amount of current is passed, then the ratio of the intensities of 
magnetic field produced at the centre of a coil will be 
(a) 9 times of first case 
(b) 
?? ?? times of first case (c) 3 times of first case 
(d) 
?? ?? times of first case 
Ans: Magnetic field at the centre of ?? turn coil carrying current ?? 
?? =
?? 0
4?? ·
2?????? ?? 
For single turn ?? = 1 
?? =
?? 0
4?? ·
2????
?? 
If the same wire is turn again to form a coil of three turns i.e. ?? = 3 and radius of each turn ?? '
=
?? 3
 
So new magnetic field at centre ?? '
=
?? 0
4?? ·
2?? (3)
?? '
 ? ?? '
= 9 ×
?? 0
4?? ·
2????
?? 
Comparing equation (ii) and (iii) gives ?? '
= 9?? . 
Q4:  An equilateral triangle of side 'a' carries a current ?? then find out the magnetic field at point ?? 
which is vertex of triangle 
(a) 
?? ?? ?? ?? v?? ????
? 
(b) 
?? ?? ?? ?? v?? ????
? 
(c) 
?? v?? ?? ?? ?? ????
?                                                                                                      
 
(d) Zero 
Ans:  
Solution: (b) As shown in the following figure magnetic field at ?? due to side 1 and side 2 is zero. 
 
Magnetic field at ?? is only due to side 3, which is ?? 1
=
?? 0
4?? ·
2?? sin 30
°
v3?? 2
? 
=
?? 0
4?? ·
2?? v3?? ?=
?? 0
?? 2v3????
? 
 
Q5:  A battery is connected between two points ?? and ?? on the circumference of a uniform 
conducting ring of radius ?? and resistance ?? . One of the arcs ???? of the ring subtends an angle ?? at 
the centre. The value of, the magnetic induction at the centre due to the current in the ring is 
(a) 
Proportional to ?? (??????
°
- ?? ) 
(b) 
Inversely proportional 
to ?? 
(c) Zero, only if ?? = ??????
°
 
(d) 
Zero for all values of 
?? 
Ans:  (d) Directions of currents in two parts are different, so directions of magnetic fields due to 
these currents are different. 
Also applying Ohm's law across ???? 
Also ?? 1
=
?? 0
4?? ×
?? 1
?? 1
?? 2
 and ?? 2
=
?? 0
4?? ×
?? 2
?? 2
?? 2
; ?
?? 2
?? 1
=
?? 1
?? 1
?? 2
?? 2
= 1  [Using (i)] 
 
Hence, two field are equal but of opposite direction. So, resultant magnetic induction at the centre is 
zero and is independent of ?? . 
Q6:  The earth's magnetic induction at a certain point is ?? × ????
-?? ???? /?? ?? . This is to be annulled 
by the magnetic induction at the centre of a circular conducting loop of radius ?? ???? . The required 
current in the loop is 
(a) ?? .???? ?? 
(b) ?? .?? ?? 
(c) ?? .???? ?? 
(d) ?? .?? ?? 
Ans:  (b) According to the question, at centre of coil ?? = ?? ?? ?
?? 0
4?? ·
2????
?? = ?? ?? ? 10
-7
×
2????
(5×1
-2
)
=
7× 10
-5
? ?? = 5.6amp . 
Q7:  A particle carrying a charge equal to 100 times the charge on an electron is rotating per 
second in a circular path of radius o. 8 metre. The value of the magnetic field produced at the 
centre will be ( ?? ?? - permeability for vacuum) 
(a) 
????
-?? ?? ?? 
(b) ????
-????
?? ?? 
(c) ????
-?? ?? ?? 
(d) ????
-?? ?? ?? 
Ans: (b) Magnetic field at the centre of orbit due to revolution of charge. 
?? =
?? 0
4?? ·
2?? (???? )
?? ; where ?? = frequency of revolution of charge 
So, ?? =
?? 0
4?? ×
2?? ×(100?? ×1)
0.8
? ?? = 10
-17
?? 0
. 
Q8:  Ratio of magnetic field at the centre of a current carrying coil of radius ?? and at a distance of 
?? ?? on its axis is 
(a) ???? v???? 
(b) ???? v???? 
(c) ?? v???? 
(d) v???? 
Page 5


Solved Example on Magnetic Effect of Electric Current 
and Magnetism 
JEE Mains 
Q1: Magnetic field due to a current carrying loop or a coil at a distant axial point ?? is ?? ?? and at an 
equal distance in it's plane is ?? ?? then 
?? ?? ?? ?? is 
(a) 2 
(b) 1 
(c) 
?? ?? 
(d) None of these 
Ans:  Current carrying coil behaves as a bar magnet as shown in figure. 
 
We also knows for a bar magnet, if axial and equatorial distance are same then ?? ?? = 2?? ?? 
Hence, in this equation 
?? 1
?? 2
=
2
1
 
Q2: Find the position of point from wire ' ?? ' where net magnetic field is zero due to following 
current distribution 
(a) ?? ???? 
(b) 
????
?? ????                                                                                           
(c) 
????
?? ???? 
(d) ?? ???? 
Ans: Suppose ?? is the point between the conductors where net magnetic field is zero. 
So at ?? | Magnetic field due to conductor 1| = | Magnetic field due to conductor 2 | 
i.e. 
?? 0
4?? ·
2(5?? )
?? =
?? 0
4?? ·
2(2?? )
(6-?? )
?
5
?? =
9
6-?? ? ?? =
30
7
 cm  
Hence position from ?? = 6 -
30
7
=
12
7
 cm 
Q3: A wire of fixed length is turned to form a coil of one turn. It is again turned to form a coil of 
three turns. If in both cases same amount of current is passed, then the ratio of the intensities of 
magnetic field produced at the centre of a coil will be 
(a) 9 times of first case 
(b) 
?? ?? times of first case (c) 3 times of first case 
(d) 
?? ?? times of first case 
Ans: Magnetic field at the centre of ?? turn coil carrying current ?? 
?? =
?? 0
4?? ·
2?????? ?? 
For single turn ?? = 1 
?? =
?? 0
4?? ·
2????
?? 
If the same wire is turn again to form a coil of three turns i.e. ?? = 3 and radius of each turn ?? '
=
?? 3
 
So new magnetic field at centre ?? '
=
?? 0
4?? ·
2?? (3)
?? '
 ? ?? '
= 9 ×
?? 0
4?? ·
2????
?? 
Comparing equation (ii) and (iii) gives ?? '
= 9?? . 
Q4:  An equilateral triangle of side 'a' carries a current ?? then find out the magnetic field at point ?? 
which is vertex of triangle 
(a) 
?? ?? ?? ?? v?? ????
? 
(b) 
?? ?? ?? ?? v?? ????
? 
(c) 
?? v?? ?? ?? ?? ????
?                                                                                                      
 
(d) Zero 
Ans:  
Solution: (b) As shown in the following figure magnetic field at ?? due to side 1 and side 2 is zero. 
 
Magnetic field at ?? is only due to side 3, which is ?? 1
=
?? 0
4?? ·
2?? sin 30
°
v3?? 2
? 
=
?? 0
4?? ·
2?? v3?? ?=
?? 0
?? 2v3????
? 
 
Q5:  A battery is connected between two points ?? and ?? on the circumference of a uniform 
conducting ring of radius ?? and resistance ?? . One of the arcs ???? of the ring subtends an angle ?? at 
the centre. The value of, the magnetic induction at the centre due to the current in the ring is 
(a) 
Proportional to ?? (??????
°
- ?? ) 
(b) 
Inversely proportional 
to ?? 
(c) Zero, only if ?? = ??????
°
 
(d) 
Zero for all values of 
?? 
Ans:  (d) Directions of currents in two parts are different, so directions of magnetic fields due to 
these currents are different. 
Also applying Ohm's law across ???? 
Also ?? 1
=
?? 0
4?? ×
?? 1
?? 1
?? 2
 and ?? 2
=
?? 0
4?? ×
?? 2
?? 2
?? 2
; ?
?? 2
?? 1
=
?? 1
?? 1
?? 2
?? 2
= 1  [Using (i)] 
 
Hence, two field are equal but of opposite direction. So, resultant magnetic induction at the centre is 
zero and is independent of ?? . 
Q6:  The earth's magnetic induction at a certain point is ?? × ????
-?? ???? /?? ?? . This is to be annulled 
by the magnetic induction at the centre of a circular conducting loop of radius ?? ???? . The required 
current in the loop is 
(a) ?? .???? ?? 
(b) ?? .?? ?? 
(c) ?? .???? ?? 
(d) ?? .?? ?? 
Ans:  (b) According to the question, at centre of coil ?? = ?? ?? ?
?? 0
4?? ·
2????
?? = ?? ?? ? 10
-7
×
2????
(5×1
-2
)
=
7× 10
-5
? ?? = 5.6amp . 
Q7:  A particle carrying a charge equal to 100 times the charge on an electron is rotating per 
second in a circular path of radius o. 8 metre. The value of the magnetic field produced at the 
centre will be ( ?? ?? - permeability for vacuum) 
(a) 
????
-?? ?? ?? 
(b) ????
-????
?? ?? 
(c) ????
-?? ?? ?? 
(d) ????
-?? ?? ?? 
Ans: (b) Magnetic field at the centre of orbit due to revolution of charge. 
?? =
?? 0
4?? ·
2?? (???? )
?? ; where ?? = frequency of revolution of charge 
So, ?? =
?? 0
4?? ×
2?? ×(100?? ×1)
0.8
? ?? = 10
-17
?? 0
. 
Q8:  Ratio of magnetic field at the centre of a current carrying coil of radius ?? and at a distance of 
?? ?? on its axis is 
(a) ???? v???? 
(b) ???? v???? 
(c) ?? v???? 
(d) v???? 
Ans:  (a) By using 
?? centre 
?? axis 
= (1+
?? 2
?? 2
)
3/2
; where ?? = 3?? and ?? = ?? ?
?? centre 
?? axis 
= (10)
3/2
= 10v10. 
Q9:  A long solenoid has 200 turns per ???? and carries a current of ?? .?? ?? . The magnetic field at its 
centre is 
[?? ?? = ?? ?? × ????
-?? ???? /?? ?? ] 
 
(a) ?? .???? × ????
-?? ???? /?? ?? 
(b) ?? .???? × ????
-?? ???? /?? ?? 
(c) ?? .???? × ????
-?? ???? /?? ?? 
(d) ???? .???? × ????
-?? ???? /?? ?? 
Ans: (b)  ?? = ?? 0
???? = 4?? × 10
-7
×
200
10
-2
× 2.5 = 6.28× 10
-2
 Wb/m
2
. 
 
Q10:  Two solenoids having lengths ?? and ?? ?? and the number of loops ?? and ?? ?? , both have the 
same current, then the ratio of the magnetic field will be 
(a) ?? :?? 
(b) ?? :?? 
(c) ?? :?? 
(d) ?? :?? 
Ans:  (a)  ?? = ?? 0
?? ?? ?? ? ?? ?
?? ?? ?
?? 1
?? 2
=
?? 1
?? 2
×
?? 2
?? 1
=
?? 4?? ×
2?? ?? =
1
2
. 
Q11: Electrons move at right angles to a magnetic field of ?? .?? × ????
-?? Tes/a with a speed of 
?? × ????
????
 ?? /?? . If the specific charge of the electron is ?? .?? × ????
????
 Coul/ ???? . The radius of the 
circular path will be 
[BHU 2003] 
(a) ?? .?? ???? 
(b) ?? .?? ???? 
(c) ?? .???? ???? 
(d) ?? ???? 
Ans: (c)  ?? =
????
????
?
?? (?? /?? ).?? =
6×10
27
17×10
11
×1.5×10
-2
= 2.35× 10
-2
 m= 2.35 cm . 
 
Q12:  An electron and a proton enter a magnetic field perpendicularly. Both have same kinetic 
energy. Which of the following is true 
(a) Trajectory of electron is less curved 
(b) Trajectory of proton is less curved 
(c) Both trajectories are equally curved 
(d) Both move on straight line path 
Ans: (b) By using ?? =
v2????
????
; For both particles ?? ? same, ?? ? same, ?? ? same