NCERT Solutions Class 12 Physics Chapter 4 Moving Charges And Magnetism
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 centre of the coil?
Answer:
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:
\(|\bar B| = \frac{\mu_{0}\; 2\pi nI}{4\pi r}\)
where μ0 is the permeability of free space = 4π × 10−7 T m A−1
hence,
\(|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\pi \times 100\times 0.4}{0.08}\)
\(= 3.14 \times 10^{-4}\; T\)
The magnitude of the magnetic field is 3.14 × 10−4 T.
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:
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:
\(|\bar B| = \frac{\mu_{0}\; 2I}{4\pi\; r}\)
where,
μ0 = Permeability of free space
= 4π × 10−7 T m A−1
Substituting the values in the equation, we get
\(|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 35}{0.2}\)
= \(3.5 \times 10^{-5}\; T\)
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 × 10−5 T.
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:
The magnitude of the current flowing in the wire is (I) = 50 A.
The 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:
\(|\bar B| = \frac{\mu_{0}\; 2I}{4\pi\; r}\)
where,
μ0 = Permeability of free space
= 4π × 10−7 T m A−1
\(|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 50}{2.5}\)
= \(4 \times 10^{-6}\; T\)
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.
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:
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, magnetic field at that point can be calculated as follows,
\(|\bar B| = \frac{\mu_{0}\; 2I}{4\pi\; r}\)
where,
μ0 = Permeability of free space
= 4π × 10−7 T m A−1
Substituting values in the above equation, we get
\(|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 90}{1.5}\)
= \(1.2 \times 10^{-5}\; T\)
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.
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:
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, θ = 30°.
The magnetic force per unit length on the wire is given as F = BIsinθ
= 0.15 × 8 × 1 × sin30°
= 0.6 N m−1
Hence, the magnetic force per unit length on the wire is 0.6 N m−1.
6. 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?
Answer:
In the problem,
the length of the wire (l) is 3 cm or 0.03 m
the magnitude of the current flowing in the wire (I) is 10 A
the strength of the magnetic field (B) is 0.27 T
the angle between the current and the magnetic field is \theta = 90°θ=90°.
the magnetic force exerted on the wire is calculated as follows:
F = BIsinθ
Substituting the values in the above equation, we get
= 0.27 × 10 × 0.03 × sin90°
= 8.1 × 10−2 N
Hence, the magnetic force on the wire is 8.1 × 10−2 N. The direction of the force can be obtained from Fleming’s left-hand rule.
7. 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.
Answer:
The magnitude of the current flowing in the wire A (IA) is 8 A
The magnitude of the current flowing in wire B (IB) is 5 A
The distance between the two wires (r) is 4 cm or 0.04 m
The length of the section of wire A (L) = 10 cm = 0.1 m
The force exerted on the length L due to the magnetic field is calculated as follows:
\(F = \frac{\mu_{o}I_{A}I_{B}L}{2\pi r}\)
where,
μ0 = Permeability of free space
= 4π × 10−7 T m A−1
Substituting the values, we get
\(F = \frac{4\pi \times 10^{-7}\times 8\times 5\times 0.1}{2\pi\times 0.04} = 2\times 10^{-5} \;N\)
The magnitude of force is 2 × 10−5N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.
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.
Answer:
Solenoid length (l) = 80 cm = 0.8 m
Five layers of windings of 400 turn each on the solenoid.
∴ Total number of turns on the solenoid, N = 5 x 400 = 2000
Solenoid Diameter (D) = 1.8 cm = 0.018 m
The current carried by the solenoid (I) = 8.0 A
The relation that gives the magnitude of the magnetic field inside the solenoid near its centre is given below:
\(B = \frac{\mu_{0}NI}{l}\)
Where,
μ0 = Permeability of free space
= 4π × 10−7 T m A−1
\(B = \frac{4\pi \times 10^{-7}\times 2000\times 8}{0.8}\)
= 2.5 × 10−2 T
Hence, the magnitude of B inside the solenoid near its centre is 2.5 × 10−2 T.