## Moving Charges and Magnetism ncert solutions: Ncert Class 12 Physics Chapter 4 Exercise Solutions

Textbook | NCERT |

Class | Class 12 |

Subject | Physics |

Chapter | Chapter 4 |

Chapter Name | Moving Charges and Magnetism class 12 exercise solutions |

Category | Ncert Solutions |

Medium | English |

Are you looking for Ncert Solution for Class 12 Physics Chapter 4 Moving Charges and Magnetism? Now you can download Ncert Class 12 Physics Chapter 4 Exercise Solutions pdf from here.

### Question 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 centre of the coil?

**Solution 4.1:** To calculate the magnitude of the magnetic field ( B ) at the center of a circular coil, we can use the formula:

\(B = \frac{\mu_0 \cdot N \cdot I}{2 \cdot R}\)

where:

- \( \mu_0 \) is the permeability of free space \(\mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}\),
- N is the number of turns of the coil,
- I is the current in the coil,
- R is the radius of the coil.

Given:

- N = 100 turns,
- I = 0.40 A,
- R = 8.0 cm = 0.080 m.

Now, let’s plug in the values:

\(B = \frac{4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \times 100 \times 0.40 \, \text{A}}{2 \times 0.080 \, \text{m}}\)

Let’s calculate this step by step.

Multiply \( \mu_0 \), N, and I:

\(4\pi \times 10^{-7} \times 100 \times 0.40 \)

\(= 5.0265 \times 10^{-5} \, \text{T} \cdot \text{m}\)

1. Multiply 2 × r:

2 × 0.080 = 0.160m

2. Divide the results from step 1 by step 2:

\(B = \frac{5.0265 \times 10^{-5}}{0.160} \, \text{T} = 3.1416 \times 10^{-4} \, \text{T}\)

So, the magnitude of the magnetic field ( B ) at the center of the coil is:

\(B = 3.14 \times 10^{-4} \, \text{T} \, \text{or} \, 0.314 \, \text{mT}\)

### Question 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?

**Solution 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:

\(|\bar B| = \frac{\mu_{0}\; 2I}{4\pi\; r}\)

where, \(\mu_{0}\) = Permeability of free space

= \(4\pi \times 10^{-7}\; T\;m\;A^{-1}\)

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 \(3.5 \times 10^{-5}\; T\)

### Question 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.

**Solution 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:

\(|\bar B| = \frac{\mu_{0}\; 2I}{4\pi\; r}\)

where, \(\mu_{0}\) = Permeability of free space

= \(4\pi \times 10^{-7}\; T\;m\;A^{-1}\)

\(|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 50}{2.5}\)

= \(4 \times 10^{-6}\; T\)

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.

### Question 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?

**Solution 4.4:** Given:

- Current in the wire, I = 90 A
- Distance from the wire, r = 1.5 m
- Magnetic constant, \( \mu_0 = 4\pi \times 10^{-7} \ \text{T m/A} \)

**Step 1: Calculate the magnitude of the magnetic field**

The magnitude of the magnetic field ( B ) at a distance ( r ) from a long straight wire carrying current ( I ) is given by:

\(B = \frac{\mu_0 I}{2\pi r}\)

Substituting the values:

\(B = \frac{(4\pi \times 10^{-7} \ \text{T m/A}) \times 90 \ \text{A}}{2\pi \times 1.5 \ \text{m}}\)

Simplifying the expression:

\(B = \frac{4\pi \times 10^{-7} \times 90}{3\pi} \ \text{T}\)

\(B = \frac{360\pi \times 10^{-7}}{3\pi} \)

\(\ \text{T} = \frac{360 \times 10^{-7}}{3} \ \text{T} = 120 \times 10^{-7} \ \text{T}\)

\(B = 1.2 \times 10^{-5} \ \text{T}\)

So, the magnitude of the magnetic field is \( 1.2 \times 10^{-5} \ \text{T} \).

**Step 2: Determine the direction of the magnetic field**

Using the right-hand rule for magnetic fields around a current-carrying wire:

- Point your thumb in the direction of the current (east-to-west in this case).
- The curl of your fingers shows the direction of the magnetic field.

Since the point where we’re calculating the magnetic field is below the wire:

- The magnetic field will point
**south**.

**Final Answer:**

The magnitude of the magnetic field 1.5 m below the wire is \( 1.2 \times 10^{-5} \ \text{T} \), and the direction is **towards the south**.

### Question 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?

**Solution 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, θ = 30°.

The magnetic force per unit length on the wire is given as F = \(BI sin\theta\)

= 0.15 × 8 × 1 × sin30°

= \(0.6\; N\; m^{-1}\)

Hence, the magnetic force per unit length on the wire is \(0.6\; N\; m^{-1}\)

### Question 4.6: 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?

**Solution 4.6:** In the problem,

- F is the magnetic force on the wire,
- I is the current through the wire,
- L is the length of the wire,
- B is the magnetic field strength,
- θ is the angle between the wire and the direction of the magnetic field.

Given:

- I = 10 A
- L = 3.0 cm = 0.03 m,
- B = 0.27 T,
- Since the wire is perpendicular to the magnetic field, \( \theta = 90^\circ \).

To calculate the magnetic force on the wire, we can use the formula:

\(F = I \cdot L \cdot B \cdot \sin(\theta)\)

Now, substituting the values into the formula:

= \(10\times 0.03\times 0.27\times sin90°\)

= \(8.1\times 10^{-2}\; N\)

Hence, the magnetic force on the wire is \(8.1\times 10^{-2}\; N\) The direction of the force can be obtained from Fleming’s left-hand rule.

### Question 4.7: 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.

**Solution 4.7:** To calculate the force between two parallel current-carrying wires, we can use the following formula:

\(F = \frac{\mu_0 I_1 I_2}{2 \pi d} \cdot L\)

where:

- F is the force between the wires,
- \(\mu_0 \) is the permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2 \),
- \( I_1 \) and \( I_2 \) are the currents in the two wires,
- d is the distance between the wires,
- L is the length of the section of the wire being considered.

Given:

- \( I_1 = 8.0 \, \text{A} \) (current in wire A),
- \( I_2 = 5.0 \, \text{A} \) (current in wire B),
- d = 4.0 cm = 0.04 m (distance between the wires),
- L = 10 cm = 0.1 m (length of the wire section).

Substituting the values into the formula:

\(F = \frac{(4\pi \times 10^{-7} \, \text{N/A}^2) \times 8.0 \, \text{A} \times 5.0 \, \text{A}}{2 \pi \times 0.04 \, \text{m}} \times 0.1 \, \text{m}\)

Simplifying:

\(F = \frac{4 \times 10^{-7} \times 8.0 \times 5.0}{0.08} \times 0.1 \, \text{N}\)

\(F = \frac{160 \times 10^{-7}}{0.08} \times 0.1 \, \text{N}\)

\(F = \frac{160 \times 10^{-7}}{8 \times 10^{-2}} \times 0.1 \, \text{N}\)

\(F = 200 \times 10^{-7} \, \text{N} \times 0.1\)

\(F = 2 \times 10^{-5} \, \text{N}\)

So the force on a 10 cm section of wire A due to the magnetic field created by wire B is \( 2.0 \times 10^{-5} \, \text{N} \).

**Direction of the Force:**

Since the currents in both wires are in the same direction, the force between them will be attractive according to the right-hand rule (for magnetic fields). This means that wire A will experience a force pulling it toward wire B.

### Question 4.8: 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.

**Solution 4.8:**

- Solenoid length (l) = 80 cm = 0.8 m
- Five layers of windings of 400 turns 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, \(\mu_{0}\) = Permeability of free space

= \(4\pi \times 10^{-7}\; T\;m\;A^{-1}\)

\(B = \frac{4\pi \times 10^{-7}\times 2000\times 8}{0.8}\)

= \(8π\times 10^{-3}\; T\) = \(2.5\times 10^{-2}\; T\)

Hence, the magnitude of B inside the solenoid near its centre is \(2.5\times 10^{-2}\; T\)

### Question 4.9: 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 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

**Solution 4.9:** The torque \(\tau\) experienced by a current-carrying coil in a magnetic field is given by the formula:

\(\tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta)\)

where:

- n is the number of turns in the coil,
- I is the current through the coil,
- A is the area of the coil,
- B is the magnetic field strength,
- θ is the angle between the normal to the plane of the coil and the magnetic field.

Given:

- n = 20 turns,
- I = 12 A,
- The side of the square coil \( l = 10 \, \text{cm} = 0.10 \, \text{m} \),
- B = 0.80 T,
- \( \theta = 30^\circ \).

**Step 1: Calculate the Area of the Coil ( A )**

The area ( A ) of the square coil is:

\(A = l^2 = (0.10 \, \text{m})^2 = 0.01 \, \text{m}^2\)

**Step 2: Calculate the Torque \(( \tau ))**

Substituting the given values into the formula:

\(\tau = 20 \times 12 \, \text{A} \times 0.01 \, \text{m}^2 \times 0.80 \, \text{T} \times \sin(30^\circ)\)

\(\tau = 20 \times 12 \times 0.01 \times 0.80 \times 0.5 \, \text{N} \cdot \text{m}\)

\(\tau = 20 \times 12 \times 0.004 \, \text{N} \cdot \text{m}\)

\(\tau = 0.96 \, \text{N} \text{m}\)

**Final Answer:**

The magnitude of the torque experienced by the coil is **0.96 N m**.

### Question 4.10: Two moving coil meters, M_{1} and M_{2} have the following particulars:

- R
_{1}= 10 Ω, N_{1}= 30, - A
_{1}= 3.6 × 10^{–3}m^{2}, B_{1}= 0.25 T - R
_{2}= 14 Ω, N_{2}= 42, - A
_{2}= 1.8 × 10^{–3}m^{2}, B_{2}= 0.50 T

(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M_{2 }and M_{1}.

**Solution 4.10:**

- For moving coil meter M
_{1} - Resistance of wire, R
_{1}= 10Ω - Number of turns, N
_{1}= 30 - Area of cross-section, A
_{1}= 3.6 × 10^{-3}m^{2} - Magnetic field strength, B
_{1}= 0.25 T - For moving coil meter M
_{2} - Resistance of wire, R
_{2}= 14Ω - Number of turns, N
_{2}= 42 - Area of cross-section, A
_{2}= 1.8 × 10^{-3}m^{2} - Magnetic field strength, B
_{2}= 0.50 T - Spring constant, K
_{1}= K_{2}= K

**(a) **Current sensitivity of M_{1} is given as: \(I_{s1} = \frac{N_{1}B_{1}A_{1}}{K_{1}}\)

And, Current sensitivity of M_{2} is given as: \(I_{s2} = \frac{N_{2}B_{2}A_{2}}{K_{2}}\)

\(∴ Ratio \frac{I_{s2}}{I_{s1}} = \frac{N_{2}B_{2}A_{2}}{N_{1}B_{1}A_{1}}\)

\(= \frac{42\times 0.5\times 1.8\times 10^{-3}\times K}{K\times 30\times 0.25\times 3.6\times 10^{-3}} = 1.4\)

Hence, the ratio of the current sensitivity of M_{2} to M_{1} is 1.4.

**(b) **Voltage sensitivity for M_{2} is given as: \(V_{s2} = \frac{N_{2}B_{2}A_{2}}{K_{2}R_{2}}\)

And voltage sensitivity for M_{1} is given as: \(V_{s1} = \frac{N_{1}B_{1}A_{1}}{K_{1}R_{1}}\)

\(∴ Ratio \frac{I_{s2}}{I_{s1}} = \frac{N_{2}B_{2}A_{2}K_{1}R_{1}}{N_{1}B_{1}A_{1}K_{2}R_{2}}\)

\(= \frac{42\times 0.5\times 1.8\times 10^{-3}\times 10\times K}{K\times 14\times 30\times 0.25\times 3.6\times 10^{-3}} = 1\)

Hence, the ratio of voltage sensitivity of M_{2} to M_{1} is 1.

### Question 4.11: In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^{-4} T) is maintained. An electron is shot into the field with a speed of 4.8 x 10^{6} m s^{-1} 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^{–19} C, m_{e}= 9.1×10^{–31} kg)

**Solution 4.11:**

- Magnetic field strength (B) = 6.5 G = \(6.5\times 10^{-4}\; T\)
- Speed of the electron (v) = \(4.8\times 10^{6}\; m/s\)
- Charge on the electron (e) = \(1.5\times 10^{-19}\; C\)
- Mass of the electron (m
_{e}) = \(9.1\times 10^{-31}\; kg\) - The angle between the shot electron and magnetic field, \(\theta = 90°\)
- The relation for Magnetic force exerted on the electron in the magnetic field is given as: \(F = evB\; sin\theta\)

This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.

Hence, the centripetal force exerted on the electron,

\(F_{e} = \frac{mv^{2}}{r}\)

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force, i.e.,

F_{e} = F

= \(\frac{mv^{2}}{r} = evB\; sin\theta\)

= \(r = \frac{mv}{eB\;sin\theta}\)

So, \(r = \frac{9.1\times 10^{-31}\times 4.8\times 10^{6}}{6.5\times 10^{-4}\times 1.5\times 10^{-19}\times sin90°}\)

\(= 4.2\times 10^{-2}\; m = 4.2 cm\)

Hence, 4.2 cm is the radius of the circular orbit of the electron.

### Question 4.12: In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

**Solution 4.12:**

- Magnetic field strength (B) = 6.5 × 10
^{–4}T - Charge on the electron (e) = 1.6 × 10
^{–19}C - Mass of the electron (m
_{e}) = 9.1 × 10^{–31}Kg - Speed of the electron (v) = 4.8 × 10
^{6}m/s - Radius of the orbit, r = 4.2 cm = 0.042 m
- Frequency of revolution of the electron = v
- Angular frequency of the electron = ω = 2πv
- The velocity of the electron is related to the angular frequency as: V = rω

Where,

- V = velocity of electron
- r = radius of path
- ω = angular frequency

In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:

\(\frac{mv^{2}}{r} = evB\)

= \(eB = \frac{mv}{r} = \frac{m(r\omega)}{r} = \frac{m(r.2\pi v)}{r}\)

= \(v = \frac{Be}{2\pi m}\)

This expression for frequency is independent of the speed of the electron. On substituting the known values in this expression, we get the frequency as:

\(v = \frac{6.5\times 10^{-4}\times 1.6\times 10^{-19}}{2\times 3.14\times 9.1\times 10^{-31}}\)

\(= 1.82\times 10^{6}\; Hz \approx 18\;MHz\)

Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

### Question 4.13: (a) A circular coil having a radius of 8.0 cm, the number of turns as 30 and 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^{0} with the normal of the coil. To prevent the coil from turning, determine the magnitude of the counter-torque that must be applied. (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.)

**Solution 4.13:**

**(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 = πr
^{2 }= π(0.08)^{2}= 0.0201 m^{2} - Current flowing in the coil (I) = 6.0 A
- Magnetic field strength, B = 1 T
- The angle between the field lines and normal with the coil surface, θ = 60°

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,

\(\tau = nIBA\; sin\theta\)

= 30 × 6 × 1 × 0.0201 × sin60°

= 3.133 N m

(b) It can be inferred from the relation \(\tau = nIBA\; sin\theta\) 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.