JEE Main Previous Year Questions (2025): Moving Charges and Magnetism

 Page 1


JEE Main Previous Year Questions 
(2025): Moving Charges and 
Magnetism 
Q1: Two long parallel wires ?? and ?? , separated by a distance of ?? ???? , carry currents 
of 5 A and 4 A , respectively, in opposite directions as shown in the figure. Magnitude 
of the resultant magnetic field at point ?? at a distance of 4 cm from wire ?? is ?? ×
????
-?? ?? . The value of ?? is ____ . Take permeability of free space as ?? ?? = ?? ?? × ????
-?? SI 
units. 
 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 1 
Solution: 
Page 2


JEE Main Previous Year Questions 
(2025): Moving Charges and 
Magnetism 
Q1: Two long parallel wires ?? and ?? , separated by a distance of ?? ???? , carry currents 
of 5 A and 4 A , respectively, in opposite directions as shown in the figure. Magnitude 
of the resultant magnetic field at point ?? at a distance of 4 cm from wire ?? is ?? ×
????
-?? ?? . The value of ?? is ____ . Take permeability of free space as ?? ?? = ?? ?? × ????
-?? SI 
units. 
 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 1 
Solution: 
 
?? =
?? 0
(5)
2?? × .01
-
?? 0
4
2?? × 0.04
 = -
100?? 0
4?? = -100× 10
-7
 = -1× 10
-5
 T
 
Q2: A proton is moving undeflected in a region of crossed electric and magnetic fields 
at a constant speed of ?? × ????
?? ????
-?? . When the electric field is switched off, the 
proton moves along a circular path of radius 2 cm . The magnitude of electric field is 
?? × ????
?? ?? /?? . The value of ?? is . Take the mass of the proton = ?? .?? × ????
-????
 ???? . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 2 
Solution: 
Let's break down the problem step by step: 
When the proton moves undeflected in crossed electric and magnetic fields, the electric and 
magnetic forces balance each other. That is, 
???? = ?????? , 
which simplifies to 
?? = ???? . 
After the electric field is switched off, the proton moves in a circular path under the action of 
the magnetic force. The magnetic force provides the required centripetal force: 
?????? =
?? ?? 2
?? . 
Solving for the magnetic field ?? , we get: 
?? =
????
????
. 
Now substitute this expression for ?? back into the equilibrium condition: 
?? = ???? = ?? (
????
????
)=
?? ?? 2
????
. 
Plug in the given values: 
Proton mass, ?? = 1.6 × 10
-27
 kg 
Speed, ?? = 2× 10
5
 m/s 
Page 3


JEE Main Previous Year Questions 
(2025): Moving Charges and 
Magnetism 
Q1: Two long parallel wires ?? and ?? , separated by a distance of ?? ???? , carry currents 
of 5 A and 4 A , respectively, in opposite directions as shown in the figure. Magnitude 
of the resultant magnetic field at point ?? at a distance of 4 cm from wire ?? is ?? ×
????
-?? ?? . The value of ?? is ____ . Take permeability of free space as ?? ?? = ?? ?? × ????
-?? SI 
units. 
 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 1 
Solution: 
 
?? =
?? 0
(5)
2?? × .01
-
?? 0
4
2?? × 0.04
 = -
100?? 0
4?? = -100× 10
-7
 = -1× 10
-5
 T
 
Q2: A proton is moving undeflected in a region of crossed electric and magnetic fields 
at a constant speed of ?? × ????
?? ????
-?? . When the electric field is switched off, the 
proton moves along a circular path of radius 2 cm . The magnitude of electric field is 
?? × ????
?? ?? /?? . The value of ?? is . Take the mass of the proton = ?? .?? × ????
-????
 ???? . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 2 
Solution: 
Let's break down the problem step by step: 
When the proton moves undeflected in crossed electric and magnetic fields, the electric and 
magnetic forces balance each other. That is, 
???? = ?????? , 
which simplifies to 
?? = ???? . 
After the electric field is switched off, the proton moves in a circular path under the action of 
the magnetic force. The magnetic force provides the required centripetal force: 
?????? =
?? ?? 2
?? . 
Solving for the magnetic field ?? , we get: 
?? =
????
????
. 
Now substitute this expression for ?? back into the equilibrium condition: 
?? = ???? = ?? (
????
????
)=
?? ?? 2
????
. 
Plug in the given values: 
Proton mass, ?? = 1.6 × 10
-27
 kg 
Speed, ?? = 2× 10
5
 m/s 
Radius, ?? = 2 cm= 0.02 m 
Proton charge, ?? = 1.6 × 10
-19
C 
Thus, 
?? =
(1.6×10
-27
 kg)(2×10
5
 m/s)
2
(1.6×10
-19
C)(0.02 m)
. 
Calculate the numerator: 
(2× 10
5
)
2
= 4× 10
10
, 
So, (1.6 × 10
-27
)× (4× 10
10
)= 6.4× 10
-17
. 
Calculate the denominator: 
(1.6 × 10
-19
)× (0.02)= 3.2× 10
-21
. 
Now compute the electric field: 
?? =
6.4×10
-17
3.2×10
-21
= 2× 10
4
 N/C. 
The problem states that the magnitude of the electric field is ?? × 10
4
 N/C. Since we found ?? =
2× 10
4
 N/C, 
it follows that 
?? = 2. 
Q3: A current of 5 A exists in a square loop of side 
?? v?? ?? . Then the magnitude of the 
magnetic field ?? at the centre of the square loop will be ?? × ????
-?? ?? . where, value of 
?? is ____ [ Take ?? ?? = ?? ?? × ????
-?? ?? ????
-?? ]. 
JEE Main 2025 (Online) 24th January Morning Shift 
Ans: 8 
Solution: 
?? =
2v2?? 0
?? ????
 
For a square loop of side length 
?? =
1
v2
 m , 
the perpendicular distance from the center to any side is 
?? =
?? 2
=
1
2v2
 m . 
For a finite straight wire, the magnetic field at a point at distance ?? from the wire is given by the 
Biot-Savart expression 
?? side 
=
?? 0
?? 4?? ?? (sin ?? 1
+ sin ?? 2
) 
where ?? 1
 and ?? 2
 are the angles between the extended wire and the line joining the ends of the 
wire with the observation point. For one side of the square, by symmetry, the midpoint of the 
side and the center of the square give 
tan ?? =
?? /2
?? /2
= 1 ? ?? = 45
°
. 
Thus, 
sin ?? 1
= sin ?? 2
= sin 45
°
=
1
v2
. 
Then the field due to one side is 
Page 4


JEE Main Previous Year Questions 
(2025): Moving Charges and 
Magnetism 
Q1: Two long parallel wires ?? and ?? , separated by a distance of ?? ???? , carry currents 
of 5 A and 4 A , respectively, in opposite directions as shown in the figure. Magnitude 
of the resultant magnetic field at point ?? at a distance of 4 cm from wire ?? is ?? ×
????
-?? ?? . The value of ?? is ____ . Take permeability of free space as ?? ?? = ?? ?? × ????
-?? SI 
units. 
 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 1 
Solution: 
 
?? =
?? 0
(5)
2?? × .01
-
?? 0
4
2?? × 0.04
 = -
100?? 0
4?? = -100× 10
-7
 = -1× 10
-5
 T
 
Q2: A proton is moving undeflected in a region of crossed electric and magnetic fields 
at a constant speed of ?? × ????
?? ????
-?? . When the electric field is switched off, the 
proton moves along a circular path of radius 2 cm . The magnitude of electric field is 
?? × ????
?? ?? /?? . The value of ?? is . Take the mass of the proton = ?? .?? × ????
-????
 ???? . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 2 
Solution: 
Let's break down the problem step by step: 
When the proton moves undeflected in crossed electric and magnetic fields, the electric and 
magnetic forces balance each other. That is, 
???? = ?????? , 
which simplifies to 
?? = ???? . 
After the electric field is switched off, the proton moves in a circular path under the action of 
the magnetic force. The magnetic force provides the required centripetal force: 
?????? =
?? ?? 2
?? . 
Solving for the magnetic field ?? , we get: 
?? =
????
????
. 
Now substitute this expression for ?? back into the equilibrium condition: 
?? = ???? = ?? (
????
????
)=
?? ?? 2
????
. 
Plug in the given values: 
Proton mass, ?? = 1.6 × 10
-27
 kg 
Speed, ?? = 2× 10
5
 m/s 
Radius, ?? = 2 cm= 0.02 m 
Proton charge, ?? = 1.6 × 10
-19
C 
Thus, 
?? =
(1.6×10
-27
 kg)(2×10
5
 m/s)
2
(1.6×10
-19
C)(0.02 m)
. 
Calculate the numerator: 
(2× 10
5
)
2
= 4× 10
10
, 
So, (1.6 × 10
-27
)× (4× 10
10
)= 6.4× 10
-17
. 
Calculate the denominator: 
(1.6 × 10
-19
)× (0.02)= 3.2× 10
-21
. 
Now compute the electric field: 
?? =
6.4×10
-17
3.2×10
-21
= 2× 10
4
 N/C. 
The problem states that the magnitude of the electric field is ?? × 10
4
 N/C. Since we found ?? =
2× 10
4
 N/C, 
it follows that 
?? = 2. 
Q3: A current of 5 A exists in a square loop of side 
?? v?? ?? . Then the magnitude of the 
magnetic field ?? at the centre of the square loop will be ?? × ????
-?? ?? . where, value of 
?? is ____ [ Take ?? ?? = ?? ?? × ????
-?? ?? ????
-?? ]. 
JEE Main 2025 (Online) 24th January Morning Shift 
Ans: 8 
Solution: 
?? =
2v2?? 0
?? ????
 
For a square loop of side length 
?? =
1
v2
 m , 
the perpendicular distance from the center to any side is 
?? =
?? 2
=
1
2v2
 m . 
For a finite straight wire, the magnetic field at a point at distance ?? from the wire is given by the 
Biot-Savart expression 
?? side 
=
?? 0
?? 4?? ?? (sin ?? 1
+ sin ?? 2
) 
where ?? 1
 and ?? 2
 are the angles between the extended wire and the line joining the ends of the 
wire with the observation point. For one side of the square, by symmetry, the midpoint of the 
side and the center of the square give 
tan ?? =
?? /2
?? /2
= 1 ? ?? = 45
°
. 
Thus, 
sin ?? 1
= sin ?? 2
= sin 45
°
=
1
v2
. 
Then the field due to one side is 
?? side 
=
?? 0
?? 4?? (
?? 2
)
(
1
v2
+
1
v2
) =
?? 0
?? 4?? (
?? 2
)
(
2
v2
) =
?? 0
?? v2
2????
. 
Since there are 4 sides with contributions that add vectorially in the same direction at the 
center, the total magnetic field is 
?? = 4· ?? side 
=
4?? 0
?? v2
2????
=
2v2?? 0
?? ????
. 
Substitute 
?? =
1
v2
, 
to obtain 
?? =
2v2?? 0
?? ?? (
1
v2
)
=
2v2?? 0
?? v2
?? =
4?? 0
?? ?? . 
Using the provided values 
?? = 5 A  and  ?? 0
= 4?? × 10
-7
 T \ m/A, 
the magnetic field becomes 
?? =
4×(4?? ×10
-7
)×5
?? =
80?? ×10
-7
?? = 80× 10
-7
 T = 8 × 10
-6
 T. 
Thus, the value of ?? is 
?? = 8. 
Q4: A tightly wound long solenoid carries a current of 1.5 A . An electron is executing 
uniform circular motion inside the solenoid with a time period of 75 ns . The number 
of turns per metre in the solenoid is 
[Take mass of electron ?? ?? = ?? × ????
-????
 ???? , charge of electron 
|?? ?? | = ?? .?? × ????
-????
?? ,?? ?? = ?? ?? × ????
-?? ?? ?? ?? ,?? ???? = ????
-?? ?? ] 
JEE Main 2025 (Online) 24th January Evening Shift 
Ans: 250 
Solution: 
The problem involves an electron executing uniform circular motion inside a solenoid. The 
objective is to find the number of turns per meter in the solenoid. 
Given: 
Current in the solenoid, ?? = 1.5 A 
Time period of the electron's circular motion, ?? = 75 ns = 75× 10
-9
 s 
Mass of electron, ?? ?? = 9× 10
-31
 kg 
Charge of electron, |?? ?? | = 1.6 × 10
-19
C 
Permeability of free space, ?? 0
= 4?? × 10
-7
 N/A
2
 
Explanation: 
The time period ?? for a revolving charge in a magnetic field is given by: 
?? =
2????
????
 
The magnetic field ?? inside a solenoid is: 
?? = ?? 0
???? 
where ?? is the number of turns per meter. Thus, substituting ?? in the expression for ?? , we get: 
Page 5


JEE Main Previous Year Questions 
(2025): Moving Charges and 
Magnetism 
Q1: Two long parallel wires ?? and ?? , separated by a distance of ?? ???? , carry currents 
of 5 A and 4 A , respectively, in opposite directions as shown in the figure. Magnitude 
of the resultant magnetic field at point ?? at a distance of 4 cm from wire ?? is ?? ×
????
-?? ?? . The value of ?? is ____ . Take permeability of free space as ?? ?? = ?? ?? × ????
-?? SI 
units. 
 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 1 
Solution: 
 
?? =
?? 0
(5)
2?? × .01
-
?? 0
4
2?? × 0.04
 = -
100?? 0
4?? = -100× 10
-7
 = -1× 10
-5
 T
 
Q2: A proton is moving undeflected in a region of crossed electric and magnetic fields 
at a constant speed of ?? × ????
?? ????
-?? . When the electric field is switched off, the 
proton moves along a circular path of radius 2 cm . The magnitude of electric field is 
?? × ????
?? ?? /?? . The value of ?? is . Take the mass of the proton = ?? .?? × ????
-????
 ???? . 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 2 
Solution: 
Let's break down the problem step by step: 
When the proton moves undeflected in crossed electric and magnetic fields, the electric and 
magnetic forces balance each other. That is, 
???? = ?????? , 
which simplifies to 
?? = ???? . 
After the electric field is switched off, the proton moves in a circular path under the action of 
the magnetic force. The magnetic force provides the required centripetal force: 
?????? =
?? ?? 2
?? . 
Solving for the magnetic field ?? , we get: 
?? =
????
????
. 
Now substitute this expression for ?? back into the equilibrium condition: 
?? = ???? = ?? (
????
????
)=
?? ?? 2
????
. 
Plug in the given values: 
Proton mass, ?? = 1.6 × 10
-27
 kg 
Speed, ?? = 2× 10
5
 m/s 
Radius, ?? = 2 cm= 0.02 m 
Proton charge, ?? = 1.6 × 10
-19
C 
Thus, 
?? =
(1.6×10
-27
 kg)(2×10
5
 m/s)
2
(1.6×10
-19
C)(0.02 m)
. 
Calculate the numerator: 
(2× 10
5
)
2
= 4× 10
10
, 
So, (1.6 × 10
-27
)× (4× 10
10
)= 6.4× 10
-17
. 
Calculate the denominator: 
(1.6 × 10
-19
)× (0.02)= 3.2× 10
-21
. 
Now compute the electric field: 
?? =
6.4×10
-17
3.2×10
-21
= 2× 10
4
 N/C. 
The problem states that the magnitude of the electric field is ?? × 10
4
 N/C. Since we found ?? =
2× 10
4
 N/C, 
it follows that 
?? = 2. 
Q3: A current of 5 A exists in a square loop of side 
?? v?? ?? . Then the magnitude of the 
magnetic field ?? at the centre of the square loop will be ?? × ????
-?? ?? . where, value of 
?? is ____ [ Take ?? ?? = ?? ?? × ????
-?? ?? ????
-?? ]. 
JEE Main 2025 (Online) 24th January Morning Shift 
Ans: 8 
Solution: 
?? =
2v2?? 0
?? ????
 
For a square loop of side length 
?? =
1
v2
 m , 
the perpendicular distance from the center to any side is 
?? =
?? 2
=
1
2v2
 m . 
For a finite straight wire, the magnetic field at a point at distance ?? from the wire is given by the 
Biot-Savart expression 
?? side 
=
?? 0
?? 4?? ?? (sin ?? 1
+ sin ?? 2
) 
where ?? 1
 and ?? 2
 are the angles between the extended wire and the line joining the ends of the 
wire with the observation point. For one side of the square, by symmetry, the midpoint of the 
side and the center of the square give 
tan ?? =
?? /2
?? /2
= 1 ? ?? = 45
°
. 
Thus, 
sin ?? 1
= sin ?? 2
= sin 45
°
=
1
v2
. 
Then the field due to one side is 
?? side 
=
?? 0
?? 4?? (
?? 2
)
(
1
v2
+
1
v2
) =
?? 0
?? 4?? (
?? 2
)
(
2
v2
) =
?? 0
?? v2
2????
. 
Since there are 4 sides with contributions that add vectorially in the same direction at the 
center, the total magnetic field is 
?? = 4· ?? side 
=
4?? 0
?? v2
2????
=
2v2?? 0
?? ????
. 
Substitute 
?? =
1
v2
, 
to obtain 
?? =
2v2?? 0
?? ?? (
1
v2
)
=
2v2?? 0
?? v2
?? =
4?? 0
?? ?? . 
Using the provided values 
?? = 5 A  and  ?? 0
= 4?? × 10
-7
 T \ m/A, 
the magnetic field becomes 
?? =
4×(4?? ×10
-7
)×5
?? =
80?? ×10
-7
?? = 80× 10
-7
 T = 8 × 10
-6
 T. 
Thus, the value of ?? is 
?? = 8. 
Q4: A tightly wound long solenoid carries a current of 1.5 A . An electron is executing 
uniform circular motion inside the solenoid with a time period of 75 ns . The number 
of turns per metre in the solenoid is 
[Take mass of electron ?? ?? = ?? × ????
-????
 ???? , charge of electron 
|?? ?? | = ?? .?? × ????
-????
?? ,?? ?? = ?? ?? × ????
-?? ?? ?? ?? ,?? ???? = ????
-?? ?? ] 
JEE Main 2025 (Online) 24th January Evening Shift 
Ans: 250 
Solution: 
The problem involves an electron executing uniform circular motion inside a solenoid. The 
objective is to find the number of turns per meter in the solenoid. 
Given: 
Current in the solenoid, ?? = 1.5 A 
Time period of the electron's circular motion, ?? = 75 ns = 75× 10
-9
 s 
Mass of electron, ?? ?? = 9× 10
-31
 kg 
Charge of electron, |?? ?? | = 1.6 × 10
-19
C 
Permeability of free space, ?? 0
= 4?? × 10
-7
 N/A
2
 
Explanation: 
The time period ?? for a revolving charge in a magnetic field is given by: 
?? =
2????
????
 
The magnetic field ?? inside a solenoid is: 
?? = ?? 0
???? 
where ?? is the number of turns per meter. Thus, substituting ?? in the expression for ?? , we get: 
?? =
2????
?? (?? 0
???? )
 
Plugging in the known values: 
75× 10
-9
=
(2?? )(9× 10
-31
)
1.6× 10
-19
× 4?? × 10
-7
× ?? × 1.5
 
Solving for ?? , we find: 
?? = 250 
Thus, the number of turns per meter in the solenoid is 250 . 
Q5: The magnetic field inside a 200 turns solenoid of radius 10 cm is ?? .?? ×
????
-?? ?????????? . If the solenoid carries a current of 0.29 A , then the length of the solenoid 
is ____ ?? ???? . 
JEE Main 2025 (Online) 29th January Evening Shift 
Ans: 8 
Solution: 
We know, magnetic field due to solenoid, ?? = ?? 0
???? 
where, I = current, ?? =
?? ?? = no. of turns per unit length 
? ?? =
?? 0
????
?? 
? ?? =
?? 0
????
?? =
4?? × 10
-7
× 200× 0.29
2.9× 10
-4
 
? ?? = 8?? × 10
-2
?? 
? ?? = 8?? cm 
Hence, answer = 8. 
Q6: A 4.0 cm long straight wire carrying a current of 8 A is placed perpendicular to a 
uniform magnetic field of strength ?? .???? ?? . The magnetic force on the wire is ____ ???? . 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 48 
Solution: 
To calculate the magnetic force on a wire, we use the formula: 
F = IlB 
Where: 
I is the current in the wire (in amperes), 
l is the length of the wire (in meters), 
B is the magnetic field strength (in teslas). 
Given that: 
The current ?? is 8 A , 
The length of the wire l is 4.0 cm , which is 0.04 m (since 1 cm= 0.01 m ), 
The magnetic field strength B is 0.15 T , 
Substituting these values into the formula gives: 
F = 8 × 0.04× 0.15