NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges and Magnetism
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1. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40A. What is the magnitude of the magnetic field B at the centre of the coil?
Sol. Here, n = 100, r = 8 cm = 8 × 10–2 m
and I = 0.40A
The magnetic field B at the centre
$$\text{B}=\frac{\mu_{0}\text{nI}}{2\text{r}}\\=\frac{4\pi×10^{\normalsize-7}×100×0.4}{2×8×10^{\normalsize-2}}\\=3.1×10^{\normalsize-4}\text{T}$$
2. A long straight wire carries a current of 35A. What is the magnitude of the field B at a point 20 cm away from the wire?
Sol. Given I = 35A and r = 20cm = 0.2m
The wire is long and it is considered as an infinite length wire. The magnetic field at point P,
$$\text{B}=\frac{\mu_{0}}{4\pi}.\frac{2\text{I}}{r}=\frac{10^{\normalsize-7}×2×35}{0.2}$$
= 3.5 × 10–5 T
3. A long straight wire in the horizontal plane carries a current of 50A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Sol. Consider a wire which is placed in north-south direction and in the horizontal plane and point P is in the east direction from the wire.
The magnitude of magnetic field
$$\text{B}=\frac{\mu_{0}}{4\pi}.\frac{2\text{I}}{r}=10^{\normalsize-7}.\frac{2×50}{2.5}$$
= 4 × 10–6 T
The direction of magnetic field at point P is perpendicularly outwards to the plane of paper.
5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8A and making an angle of 30° with the direction of a uniform magnetic field of 0.15T?
Sol. Here I = 8A, q = 30°, B = 0.15T, l = 1m
The magnitude of magnetic force
F = I(l × B) = IlB sin θ
= 8 × 1 × 0.15 × sin 30°
$$=\frac{8×0.15}{2}=0.6\text{N/m}$$
The direction of force is perpendicular to both the direction of magnetic field and the direction of flow of current both, the direction of force is perpendicularly inwards to the plane of paper.
6. A 3.0 cm wire carrying a current of 10A 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?
Sol. Given, l = 3 cm = 3 × 10–2m, I = 10A, B = 0.2 T.
The magnitude of magnetic force on the wire
F = IlB sin 90° = 10 × 3 × 10–2 × 0.27 × sin 90° = 8.1 × 10–2N
According to right hand palm rule, the direction of magnetic force is perpendicular to plane of paper inwards.
7. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A?
Sol. Force per unit length on two parallel wire carrying current.
$$\text{F}=\frac{\mu_{0}}{4\pi}.\frac{2\text{I}_{1}.\text{I}_{2}}{r}\\=\frac{10^{\normalsize-7}×2×8×5}{0.04}=2×10^{\normalsize-4}\text{N}$$
The force on A of length 10 cm is F’ = F × 0.1
(∵ 1m = 100 cm)
⇒ F’ = 2 ×10–4 × 0.1 = 2 × 10–5N
According to Maxwell’s right hand rule the direction of magnetic field due to B on A is perpendicularly outwards to the plane of paper.
According to Fleming’s left hand rule, the direction of force is towards B.
8. 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.
Sol. The length of solenoid, l = 80 cm = 0.8 m
Number of layers = 5
Number of turns per layer = 400
∴ The total number of turns N = 400 × 5 = 2000
Diameter of solenoid = 1.8 cm
Current in solenoid I = 8A
$$\text{and number of turns / length, n =}\frac{2000}{0.8}=2500$$
The magnitude of magnetic field inside the solenoid
B = µ0nI
= 4 × 3.14 × 10–7 × 2500 × 8
= 2.5 × 10–2T
The direction of magnetic field is along the axis of solenoid.
9. A square coil of side 10 cm consists of 20 turns and carries a current of 12A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Sol. Here, side of square coil = 10 cm = 0.1m
Number of turns (n) = 20
Current in square coil I = 12A
Angle made by coil θ = 30°
Magnetic field B = 0.80T
The torque experienced by the coil
t = NI AB sin θ
= 20 × 12 × (10 × 10–2)2 × 0.80 × sin 30°
$$τ=2.4×0.80\text{sin 30}\degree=\frac{2.4×0.80}{2}$$
= 0.96 N-m
10. Two moving coil meters M1 and M2 having the following particulars:
R1 = 10Ω, N1 = 30, A1 = 3.6 × 10–3 m2, B1 = 0.25T
R2 = 14Ω, N2 = 42, A2 = 1.8 × 10–3m2, B2 = 0.50T
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.
Sol. Given R1 = 10Ω, N1 = 30, A1 = 3.6 × 10–3 m2, B1 = 0.25T
R2 = 14Ω, N2 = 42, A2 = 1.8 × 10–3m2, B2= 0.50T
k1 = k2
(a) The current sensitivity,
$$\text{I}=\frac{\text{NAB}}{k}\\\therefore\space\frac{\text{Is}_{2}}{\text{Is}_{1}}-\frac{\text{N}_{2}\text{B}_{2}\text{A}_{2}.k_{1}}{\text{N}_{1}\text{B}_{1}\text{A}_{1}k_{2}}\\=\frac{42×0.50×1.8×10^{\normalsize-3}}{30×0.25×3.6×10^{\normalsize-3}}$$
= 1.4
(b) The voltage sensitivity,
$$\text{V}=\frac{\text{NAB}}{\text{kR}}\\\frac{\text{Vs}_{2}}{\text{Vs}_{1}}=\frac{\text{N}_{2}\text{B}_{2}\text{A}_{2}.k_{1}\text{R}_{1}}{k_{2}.\text{R}_{2}.\text{N}_{1}\text{B}_{1}\text{A}_{1}}\\=\frac{42×0.50×1.8×10^{\normalsize-3}×k×10}{k×14×30×0.25×3.6×10^{\normalsize-3}}=1$$
11. In a chamber, a uniform magnetic field of 6.5 G (1G = 10–4T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m/s normal to the field explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.5 × 10–19C, me = 9.1 × 10–31 kg)
Sol. Given, magnetic field
B = 6.5 G = 6.5 × 10–4T
Charge e = – 1.6 × 10–19C
Speed of electron v = 4.8 × 106 m/s
Mass of electron me = 9.1 × 10–31 kg
Angle between magnetic field and path of electron θ = 90°.
The force on the charge particle entering in the magnetic field
F = q (v × B) = e (v × B)
According to the right hand palm rule, the direction of force is perpendicular to both v and B.
So, the force will only change the direction of motion but not the magnitude of velocity. So, the electron attains a circular path and the necessary centripetal force is provided by the magnetic force.
$$\text{e(v×B)}=\frac{\text{mv}^{2}}{r}\\\text{or evB sin 90°}=\frac{\text{mv}^{2}}{r}\\\therefore\space\text{r}=\frac{\text{mv}}{\text{eB}×1}=\frac{9.1×10^{-31}×4.8×10^{6}}{1.6×10^{-19}×6.5×10^{-4}}$$
= 4.2 × 10–2m = 4.2cm
12. In Q. 11. obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Sol. Given, B = 6.5 G = 6.5 × 10–4 T
v = 4.8 × 106 m/s, e = 1.6 × 10–19C
and me = 9.1 × 10–31kg
We know that when an electron moves on a circular path in uniform magnetic field, then the required centripetal force is provided by the magnetic force on it.
$$\frac{\text{mv}^{2}}{r}=\text{qvB}\\\Rarr\space\frac{\text{mv}}{r}=\text{qB}$$
If angular velocity of electron is w, then
v = rω
$$\therefore\space\frac{m(r\omega)}{r}=\text{qB}\\\omega=\frac{\text{qB}}{m}\space(\because \omega=2\pi n)\\\therefore\space2\pi n=\frac{\text{qB}}{m}\Rarr n= \frac{\text{qB}}{2\pi m}$$
Frequency of revolution of electron
$$\text{v}=\frac{\text{Bq}}{2\pi m}=\frac{\text{Bq}}{2\pi m_{e}}\\=\frac{6.5×10^{\normalsize-4}×1.6×10^{\normalsize-19}}{2×3.1 4×9.1×10^{-31}}$$
(∵ For electron q = e)
= 18.18 × 106 Hz
Here, we observe that the frequency of electron is independent to the velocity.
13. (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 60° 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 you 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)
Sol. (a) Given, number of turns n = 30, radius (r) = 8 cm = 0.08 m
Current in the coil I = 6A
Magnetic field B = 1.0T
Angle made by field with the normal of the coil, q = 60°
Magnitude of torque acting on the current
Carrying coil
t = nIAB sin q
= 30 × 6 × p (0.08)2 × 1 × sin 60°
$$\text{= 30 × 6 × 3.14 × 0.08 × 0.08 ×}\sqrt{\frac{3}{2}}$$
= 3.133 N-m
(b) From the formula, it is clear that the torque on the loop does not depend on the shape if area remains constant. So, the torque remains constant.
