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NCERT Class 12 Physics Textbook Chapter 4 With Answer PDF Free Download
Chapter 4: Moving Charges and Magnetism
4.1 INTRODUCTION
Both Electricity and Magnetism have been known for more than 2000 years. However, it was only about 200 years ago, in 1820, that it was realized that they were intimately related.
During a lecture demonstration in the summer of 1820, the Danish physicist Hans Christian Oersted noticed that a current in a straight wire caused a noticeable deflection in a nearby magnetic compass needle.
He investigated this phenomenon. He found that the alignment of the needle is tangential to an imaginary circle which has the straight wire as its center and has its plane perpendicular to the wire.
This situation is depicted in Fig.4.1(a). It is noticeable when the current is large and the needle sufficiently close to the wire so that the earth’s magnetic field may be ignored.
Reversing the direction of the current reverses the orientation of the needle [Fig. 4.1(b)].
The deflection increases by increasing the current or bringing the needle closer to the wire. Iron filings sprinkled around the wire arrange themselves in concentric circles with the wire as the center [Fig. 4.1(c)].
Oersted concluded that moving charges or currents produced a magnetic field in the surrounding space.
Magnetic force on a current-carrying conductor
We can extend the analysis for force due to the magnetic field on a single moving charge to a straight rod carrying current.
Consider a rod of a uniform cross-sectional area A and length l. We shall assume one kind of mobile carrier as a conductor (here electrons).
Let the number density of these mobile charge carriers in it be n. Then the total number of mobile charge carriers in it is nlA.
For a steady current I in this conducting rod, we may assume that each mobile carrier has an average drift velocity vd (see Chapter 3).
In the presence of an external magnetic field B, the force on these carriers is:
F = (nlA)q vd ×× B
where q is the value of the charge on a carrier. Now nqvd is the current density j and |(nq vd )|A is the current I (see Chapter 3 for the discussion of current and current density). Thus,
F = [(nqvd )l A] × B = [ jAl ] × B
= Il × B (4.4)
where l is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector.
In the last step leading to Eq. (4.4), we have transferred the vector sign from j to I. Equation (4.4) holds for a straight rod. In this equation, B is the external magnetic field. It is not the field produced by the current-carrying rod.
If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips dl j and summing
F B =∑Id × l
This summation can be converted to an integral in most cases.
Motion In A Magnetic Field
We will now consider, in greater detail, the motion of a charge moving in a magnetic field.
We have learned in Mechanics (see Class XI book, Chapter 6) that a force on a particle does work if the force has a component along (or opposed to) the direction of motion of the particle.
In the case of the motion of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle.
So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed).
[Notice that this is unlike the force due to an electric field, qE, which can have a component parallel (or antiparallel) to motion and thus can transfer energy in addition to momentum.]
We shall consider the motion of a charged particle in a uniform magnetic field. First, consider the case of v perpendicular to B.
The perpendicular force, q v × B, acts as a centripetal force and produces a circular motion perpendicular to the magnetic field.
The particle will describe a circle if v and B are perpendicular to each other (Fig. 4.5).
| Author | NCERT |
| Language | English |
| No. of Pages | 41 |
| PDF Size | 13.1 MB |
| Category | Egnlish Book |
| Source/Credits | ncert.nic.in |
NCERT Solutions Class 12 Physics Chapter 4 Moving Charges and Magnetism
Q 4.1) 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 center of the coil?
Answer 4.1:
Given:
The number of turns on the coil (n) is 100
The radius of each turn (r) is 8 cm or 0.08 m
The magnitude of the current flowing in the coil (I) is 0.4 A
The magnitude of the magnetic field at the centre of the coil can be obtained by the following relation:|\bar B| = \frac{\mu_{0}\; 2\pi nI}{4\pi r}∣Bˉ∣=4πrμ02πnI
where \mu_{0}μ0 is the permeability of free space = 4\pi \times 10^{-7}\; T\;m\;A^{-1}4π×10−7TmA−1
hence,|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\pi \times 100\times 0.4}{0.08}∣Bˉ∣=4π4π×10−7×0.082π×100×0.4
= 3.14 \times 10^{-4}\; T3.14×10−4T
The magnitude of the magnetic field is 3.14 \times 10^{-4}\; T3.14×10−4T.
Q 4.2) 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?
Answer 4.2:
The magnitude of the current flowing in the wire (I) is 35 A
The distance of the point from the wire (r) is 20 cm or 0.2 m
At this point, the magnitude of the magnetic field is given by the relation:|\bar B| = \frac{\mu_{0}\; 2I}{4\pi\; r}∣Bˉ∣=4πrμ02I
where,\mu_{0}μ0 = Permeability of free space
= 4\pi \times 10^{-7}\; T\;m\;A^{-1}4π×10−7TmA−1
Substituting the values in the equation, we get|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 35}{0.2}∣Bˉ∣=4π4π×10−7×0.22×35
= 3.5 \times 10^{-5}\; T3.5×10−5T
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 \times 10^{-5}\; T3.5×10−5T.
Q 4.3) A long straight wire in the horizontal plane carries a current of 50 A in the north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Answer 4.3:
The magnitude of the current flowing in the wire is (I) = 50 A.
Point B is 2.5 m away from the East of the wire.
Therefore, the magnitude of the distance of the point from the wire (r) is 2.5 m
The magnitude of the magnetic field at that point is given by the relation:|\bar B| = \frac{\mu_{0}\; 2I}{4\pi\; r}∣Bˉ∣=4πrμ02I
where,\mu_{0}μ0 = Permeability of free space
= 4\pi \times 10^{-7}\; T\;m\;A^{-1}4π×10−7TmA−1|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 50}{2.5}∣Bˉ∣=4π4π×10−7×2.52×50
= 4 \times 10^{-6}\; T4×10−6T
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 Maxwell’s right-hand thumb rule, the direction of the magnetic field at the given point is vertically upward.
Q 4.4) A horizontal overhead power line carries a current of 90 A in the east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Answer 4.4:
The magnitude of the current in the power line is (I) = 90 A
The point is located below the electrical cable at distance (r) = 1.5 m
Hence, the magnetic field at that point can be calculated as follows,|\bar B| = \frac{\mu_{0}\; 2I}{4\pi\; r}∣Bˉ∣=4πrμ02I
where \mu_{0}μ0 = Permeability of free space
= 4\pi \times 10^{-7}\; T\;m\;A^{-1}4π×10−7TmA−1
Substituting values in the above equation, we get|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 90}{1.5}∣Bˉ∣=4π4π×10−7×1.52×90
= 1.2 \times 10^{-5}\; T1.2×10−5T
The current flows from East to West. The point is below the electrical cable.
Hence, according to Maxwell’s right-hand thumb rule, the direction of the magnetic field is towards the South.
Q 4.5) 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?
Answer 4.5:
In the problem,
The current flowing in the wire is (I) = 8 A
The magnitude of the uniform magnetic field (B) is 0.15 T
The angle between the wire and the magnetic field, \theta = 30°θ=30°.
The magnetic force per unit length on the wire is given as F = BI sin\thetaBIsinθ
= 0.15\times 8\times 1 \times sin 30°0.15×8×1×sin30°
= 0.6\; N\; m^{-1}0.6Nm−1
Hence, the magnetic force per unit length on the wire is 0.6\; N\; m^{-1}0.6Nm−1.
NCERT Class 12 Physics Textbook Chapter 4 With Answer PDF Free Download