CBSE Class 12 Physics Chapter-4 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 4 – Moving Charges and Magnetism 3 Marks Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter-4 Important Questions
CBSE Class 12 Physics Important Questions Chapter 4 – Moving Charges and Magnetism
3 Marks Questions
1. Derive an expression for the force acting on a current carrying conductor placed in a uniform magnetic field Name the rule which gives the direction of the force. Write the condition for which this force will have (1) maximum (2) minimum value?
Ans. A conductor is placed in a uniform magnetic field 
which makes and angle 
with . Let I current flows through the conductor.
If n is the no. of electrons per unit volume of the conductor, then Total no. of electrons in small current element d
= nAdl
=>= Ne
=> = nAdl e
be the force experienced by each electron
= e 
Force experienced by small current element
dF = neAvd dl B sin
(I = neAvd)
=> df = IdlBsin
Hence total force experienced
F = 
F = IB
In vector form 
= I 
(a) Force will be maximum when = 900
(b) Force will be minimum when = 00
2. A straight wire carries a current of 10A. An electron moving at 
is at distance 2.0 cm from the wire. Find the force acting on the electron if its velocity is directed towards the wire?
Ans. Here I = 10A
R = 2.0 cm = 2
Force acting on moving electron (F) = qVB sin
tesla and 
to the plane of paper and directed downwards.
Now 
Newton.
3. State Biot- Savarts law. Derive an expression for magnetic field at the centre of a circular coil of n-turns carrying current – I?
Ans. Biot – Savart law states that the magnetic field db due to a current element 
at any point is
ie dB 
I
dB dl
dB sin
dB 
Combining all we get
dB 
Consider a circular loop of radius r carrying a current I.
Since dl 
=> 
Applying Biot Savart law
dB = 
For entire closed circular loop
B = 
B = 
For n turns of a coil 
4. What is radial magnetic field? How it is obtained in moving coil galvanometer?
Ans. A radial magnetic field is one in which plane of the coil always lies in the direction of the magnetic field. It can be obtained by
(a) Properly cutting the pole pieces concave in shape.
(b) Placing soft iron cylindrical core between the pole pieces.
5. Two straight parallel current carrying conductors are kept at a distanced r from each other in air. The direction for current in both the conductor is same. Find the magnitude and direction of the force between them. Hence define one ampere?
Ans. Consider two parallel conductors carrying or current 
and is separated by a distance ‘d’.
Magnetic field due to current 
at any paint on conductor (2) is
(to the plane & Downwards 
)
Since current carrying conductor is placed at right angles to the magnetic field
=> F = BI l 
F = B I l
=> Force experienced per unit length of conductor —-(2)
= 
—-(2)
Fleming’s left hand Rule says is directed towards conductor (1)
Similarly 
(Directed Towards conductor (2))
Since 
are equal and opposite so two parallel current carrying conductor attract each other.
Since F = 
If
= 1A d = 1m
Thus one ampere is that current which is flowing in two infinitely long parallel conductors separated by a distance of 1 meter in vacuum and experiences a force of on each meter of the other wire.
6. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
Ans. Number of turns on the circular coil, n = 100
Radius of each turn, r = 8.0 cm = 0.08 m
Current flowing in the coil, I = 0.4 A
Magnitude of the magnetic field at the centre of the coil is given by the relation,
| B | = 
Where,
= Permeability of free space
= 4π × 10–7 T m A–1
| B |=
Hence, the magnitude of the magnetic field is .
7. A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?
Ans. Current in the wire, I = 35 A
Distance of a point from the wire, r = 20 cm = 0.2 m
Magnitude of the magnetic field at this point is given as:
B
Where,
= Permeability of free space = 
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is
8. A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Ans. Current in the wire, I = 50 A
A point is 2.5 m away from the East of the wire.
Magnitude of the distance of the point from the wire, r = 2.5 m.
Magnitude of the magnetic field at that point is given by the relation, B
Where,
= Permeability of free space = 
The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.
9. A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Ans. Current in the power line, I = 90 A
Point is located below the power line at distance, r = 1.5 m
Hence, magnetic field at that point is given by the relation,
B
Where,
= Permeability of free space =
The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell’s right hand thumb rule, the direction of the magnetic field is towards the South.
10. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?
Ans. Current in the wire, I = 8 A
Magnitude of the uniform magnetic field, B = 0.15 T
Angle between the wire and magnetic field, 
= 30°.
Magnetic force per unit length on the wire is given as:
f= BI sin
= 
=
Hence, the magnetic force per unit length on the wire is.
11. A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
Ans. Length of the wire, l = 3 cm = 0.03 m
Current flowing in the wire, I = 10 A
Magnetic field, B = 0.27 T
Angle between the current and magnetic field, = 90°
Magnetic force exerted on the wire is given as:
F= BIl sin
= 
= 
Hence, the magnetic force on the wire is . The direction of the force can be obtained from Fleming’s left hand rule.
12. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Ans. Current flowing in wire A, 
Current flowing in wire B, 
Distance between the two wires, r = 4.0 cm = 0.04 m
Length of a section of wire A, l = 10 cm = 0.1 m
Force exerted on length l due to the magnetic field is given as:
Where,
= Permeability of free space =
The magnitude of force is . This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.
13. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
Ans. Length of the solenoid, l = 80 cm = 0.8 m
There are five layers of windings of 400 turns each on the solenoid.
Total number of turns on the solenoid, N = 5 × 400 = 2000
Diameter of the solenoid, D = 1.8 cm = 0.018 m
Current carried by the solenoid, I = 8.0 A
Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,
Where,
= Permeability of free space =
Hence, the magnitude of the magnetic field inside the solenoid near its centre is
14. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 
with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Ans. Length of a side of the square coil, l = 10 cm = 0.1 m
Current flowing in the coil, I = 12 A
Number of turns on the coil, n = 20
Angle made by the plane of the coil with magnetic field, 
Strength of magnetic field, B = 0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
T = n BIA sinθ
Where,
A = Area of the square coil
= 0.96 N m
Hence, the magnitude of the torque experienced by the coil is 0.96 N m.
15. (a) 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 
with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) 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.)
Ans. (a) Number of turns on the circular coil, n = 30
Radius of the coil, r = 8.0 cm = 0.08 m
Area of the coil 
Current flowing in the coil, I = 6.0 A
Magnetic field strength, B = 1 T
Angle between the field lines and normal with the coil surface,
The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation,
… (i)
= 
= 3.133 N m
(b) It can be inferred from relation (i) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.
16. A magnetic field of 100 G 
is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about
. 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 
. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Ans. Magnetic field strength, B = 100 
Number of turns per unit length, 
Current flowing in the coil, I = 15 A
Permeability of free space, =
Magnetic field is given by the relation, 
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.
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.
Ans. Inner radius of the toroid, 
= 0.25 m
Outer radius of the toroid, 
= 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,
Where,
= Permeability of free space =
l = length of toroid
(c) Magnetic field in the empty space surrounded by the toroid is zero.
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.
Ans.(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.
19. A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.)
.
Ans. Length of the rod, l = 0.45 m
Mass suspended by the wires, m = 60 g =
Acceleration due to gravity,
Current in the rod flowing through the wire, I = 5 A
(a) Magnetic field (B) is equal and opposite to the weight of the wire i.e., BIl + mg
A horizontal magnetic field of 0.26 T normal to the length of the conductor should be set up in order to get zero tension in the wire. The magnetic field should be such that Fleming’s left hand rule gives an upward magnetic force.
(b) If the direction of the current is revered, then the force due to magnetic field and the weight of the wire acts in a vertically downward direction.
∴Total tension in the wire = BIl + mg
= 1.176 N
20. The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
Ans. Current in both wires, I = 300 A
Distance between the wires, r = 1.5 cm = 0.015 m
Length of the two wires, l = 70 cm = 0.7 m
Force between the two wires is given by the relation,
Where,
= Permeability of free space =
=1.2 N/m
Since the direction of the current in the wires is opposite, a repulsive force exists between them.
21. A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area
, and the free electron density in copper is given to be about 
.)
Ans. Number of turns on the circular coil, n = 20
Radius of the coil, r = 10 cm = 0.1 m
Magnetic field strength, B = 0.10 T
Current in the coil, I = 5.0 A
(a) The total torque on the coil is zero because the field is uniform.
(b) The total force on the coil is zero because the field is uniform.
(c) Cross-sectional area of copper coil, A =
Number of free electrons per cubic meter in copper, N =
Charge on the electron, 
Magnetic force, 
Where,
= Drift velocity of electrons
Hence, the average force on each electron is
22. galvanometer coil has a resistance of 12 
and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?
Ans. Resistance of the galvanometer coil, G = 12
Current for which there is full scale deflection, 
= 3 mA = 
Range of the voltmeter is 0, which needs to be converted to 18 V.
Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:
= 
Hence, a resistor of resistance is to be connected in series with the galvanometer.
23. A galvanometer coil has a resistance of 15 and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?
Ans. Resistance of the galvanometer coil, G = 15
Current for which the galvanometer shows full scale deflection,
= 4 mA = 
Range of the ammeter is 0, which needs to be converted to 6 A.
Current, I = 6 A
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:
Hence, a 
shunt resistor is to be connected in parallel with the galvanometer.