Chapter 4: Moving Charges and Magnetism Application Questions | physics Class 12 CBSE

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Chapter 4: Moving Charges and Magnetism Application Questions | physics Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A wire carrying 200 g mass and 1.5 m long is suspended in mid-air by a uniform horizontal magnetic field. The current through the wire is 2 A. With reference to the force balance involved, determine the magnitude of the magnetic field required.

The wire experiences an upward magnetic force F = BIl and a downward gravitational force mg. For mid-air suspension these must balance: mg = BIl, so B = mg/(Il).
Substituting values: B = (0.2 × 9.8)/(2 × 1.5) = 1.96/3 ≈ 0.65 T. The magnetic field direction must be horizontal and perpendicular to the current direction so that the force Il × B is directed upward.
Note that only m/l (mass per unit length) determines B; also, the earth's magnetic field (~4 × 10⁻⁵ T) is negligible compared to the required 0.65 T and is correctly ignored in this calculation.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

An electron moves with a speed of 3 × 10⁷ m/s in a magnetic field of 6 × 10⁻⁴ T perpendicular to its velocity. Calculate the radius of its circular path, frequency of revolution, and energy in keV.

Radius: r = mv/(qB) = (9 × 10⁻³¹ × 3 × 10⁷)/(1.6 × 10⁻¹⁹ × 6 × 10⁻⁴) = 27 × 10⁻²⁴/(9.6 × 10⁻²³) = 28 × 10⁻² m = 28 cm.
Frequency: ν = v/(2πr) = (3 × 10⁷)/(2π × 0.28) ≈ 17 × 10⁶ Hz = 17 MHz. This cyclotron frequency is independent of the electron's speed for a given B and m/q ratio.
Energy: E = (1/2)mv² = (1/2)(9 × 10⁻³¹)(9 × 10¹⁴) = 40.5 × 10⁻¹⁷ J ≈ 4 × 10⁻¹⁶ J = 2.5 keV. (Using 1 eV = 1.6 × 10⁻¹⁹ J gives ≈ 2.5 keV.)

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 3

Applied Concept

With reference to a long solenoid of length 0.5 m, radius 1 cm and 500 turns carrying 5 A, verify that the long solenoid approximation applies and calculate the magnetic field inside.

Number of turns per unit length: n = 500/0.5 = 1000 turns/m. Check of approximation: l/a = 0.5/0.01 = 50, so l >> a — the long solenoid formula B = μ₀nI is valid.
Magnetic field inside: B = μ₀nI = 4π × 10⁻⁷ × 10³ × 5 = 4π × 10⁻⁴ × 5 = 6.28 × 10⁻³ T ≈ 6.3 mT. The field is uniform along the interior axis.
The field outside is negligible (approaches zero for an ideal long solenoid). Inserting a soft iron core would significantly increase the field inside, a technique used in electromagnets.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 4

Applied Concept

A 100-turn tightly wound coil of radius 10 cm carries a current of 1 A. Referring to the principle of superposition applied to multi-turn coils, calculate the magnetic field at the centre.

Since the coil is tightly wound, all N = 100 turns have the same radius R = 0.1 m. Each turn contributes equally to the field at the centre.
By superposition, the total field is N times the field due to one turn: B = Nμ₀I/(2R) = 100 × 4π × 10⁻⁷ × 1/(2 × 0.1) = 4π × 10⁻⁵/0.2 = 2π × 10⁻⁴ T.
Numerically, B = 6.28 × 10⁻⁴ T. The direction is along the axis of the coil, determined by the right-hand thumb rule applied to the current direction.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A square coil of side 10 cm with 20 turns carries 12 A. It is suspended vertically with the normal to its plane at 30° to a horizontal field of 0.80 T. Calculate the torque and discuss whether the coil is in equilibrium.

Magnetic moment: m = NIA = 20 × 12 × (0.1)² = 20 × 12 × 0.01 = 2.4 A m². The angle between m and B is θ = 30°.
Torque: τ = mB sinθ = 2.4 × 0.80 × sin30° = 2.4 × 0.80 × 0.5 = 0.96 N m.
The coil is not in equilibrium since the torque is non-zero (equilibrium requires θ = 0° or 180°). The torque will rotate the coil toward the stable equilibrium position where m is parallel to B (θ = 0°).

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

With reference to a galvanometer circuit (3 V source, 3 Ω resistance), analyse the change in measured current when (a) a galvanometer with RG = 60 Ω is used, (b) it is converted to an ammeter with rs = 0.02 Ω, (c) an ideal ammeter is used.

Case (a): Total resistance = RG + 3 = 63 Ω; current I = 3/63 ≈ 0.048 A. The galvanometer's large resistance drastically reduces the current from its true value.
Case (b): Ammeter resistance ≈ rs = (60 × 0.02)/(60 + 0.02) ≈ 0.02 Ω; total resistance ≈ 3.02 Ω; I = 3/3.02 ≈ 0.99 A. The shunt makes the instrument nearly non-invasive.
Case (c): Ideal ammeter has zero resistance; total resistance = 3 Ω; I = 3/3 = 1.00 A. This is the true circuit current. Cases (b) and (c) show why shunt resistance must be very small for accurate current measurement.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A magnetic field B is parallel to the positive y-axis. An electron moves along the positive x-axis and a proton also moves along the positive x-axis. With reference to the Lorentz force, determine the direction of force on each particle.

The velocity v is along +x-axis and B is along +y-axis. The cross product v × B = x-hat × y-hat = z-hat (using right-hand rule for cross product).
For the proton (positive charge): F = q(v × B) = +q(+z-hat) → force is along +z-axis (out of page).
For the electron (negative charge): F = –e(v × B) = –e(+z-hat) → force is along –z-axis (into page). The forces on proton and electron are equal in magnitude but opposite in direction.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 8

Applied Concept

A 3 cm wire carrying 10 A is placed inside a solenoid perpendicular to its axis where B = 0.27 T. Calculate the magnetic force on the wire and state its direction using the right-hand rule.

The wire is perpendicular to the solenoid axis and hence perpendicular to the field B inside the solenoid (θ = 90°). Force: F = BIl sinθ = 0.27 × 10 × 0.03 × 1 = 8.1 × 10⁻² N.
F = 8.1 × 10⁻² N = 81 mN. This force is directed perpendicular to both the wire (current direction) and B, determined by the right-hand rule for the cross product Il × B.
If the current in the wire is along x-axis and B is along z-axis (solenoid axis), then F = Il × B = I(l x-hat) × (B z-hat) = IlB(x-hat × z-hat) = –IlB y-hat, so the force is along –y-axis.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

Two parallel wires A and B are separated by 4 cm and carry currents 8 A and 5 A respectively in the same direction. Estimate the force on a 10 cm section of wire A and state whether the wires attract or repel.

The magnetic field due to wire B at wire A: Ba = μ₀Ib/(2πd) = (4π × 10⁻⁷ × 5)/(2π × 0.04) = 2.5 × 10⁻⁵ T.
Force on 10 cm section of wire A: F = IaLBa = 8 × 0.1 × 2.5 × 10⁻⁵ = 2 × 10⁻⁵ N. Alternatively, using f = μ₀IaIb/(2πd): f = (4π × 10⁻⁷ × 8 × 5)/(2π × 0.04) = 2 × 10⁻⁴ N/m; F = 2 × 10⁻⁴ × 0.1 = 2 × 10⁻⁵ N.
Since the currents are in the same direction (parallel), the wires attract each other. The force on A is toward B and the force on B is toward A, consistent with Newton's third law.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

A horizontal wire carries 1 A in the east-to-west direction in a region where the horizontal component of the earth's field is 3 × 10⁻⁵ T directed from south to north. Find the force per unit length and its direction.

Current direction: east to west (–x direction). Earth's horizontal field B: south to north (+y direction). The angle between the current direction (–x) and B (+y) is 90°.
Force per unit length: f = IB sinθ = 1 × 3 × 10⁻⁵ × sin90° = 3 × 10⁻⁵ N/m. This is larger than the 2 × 10⁻⁷ N/m quoted in the definition of the ampere, which is why the earth's field must be eliminated during standardisation.
Direction: F = Il × B = I(–x-hat) × (B y-hat) = –IB(x-hat × y-hat) = –IB z-hat, i.e., the force is directed downward (–z), pressing the wire toward the ground.

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Chapter 4: Moving Charges and Magnetism | Application Questions

A wire carrying 200 g mass and 1.5 m long is suspended in mid-air by a uniform horizontal magnetic field. The current through the wire is 2 A. With reference to the force balance involved, determine the magnitude of the magnetic field required.

An electron moves with a speed of 3 × 10⁷ m/s in a magnetic field of 6 × 10⁻⁴ T perpendicular to its velocity. Calculate the radius of its circular path, frequency of revolution, and energy in keV.

With reference to a long solenoid of length 0.5 m, radius 1 cm and 500 turns carrying 5 A, verify that the long solenoid approximation applies and calculate the magnetic field inside.

A 100-turn tightly wound coil of radius 10 cm carries a current of 1 A. Referring to the principle of superposition applied to multi-turn coils, calculate the magnetic field at the centre.

A square coil of side 10 cm with 20 turns carries 12 A. It is suspended vertically with the normal to its plane at 30° to a horizontal field of 0.80 T. Calculate the torque and discuss whether the coil is in equilibrium.

With reference to a galvanometer circuit (3 V source, 3 Ω resistance), analyse the change in measured current when (a) a galvanometer with RG = 60 Ω is used, (b) it is converted to an ammeter with rs = 0.02 Ω, (c) an ideal ammeter is used.

A magnetic field B is parallel to the positive y-axis. An electron moves along the positive x-axis and a proton also moves along the positive x-axis. With reference to the Lorentz force, determine the direction of force on each particle.

A 3 cm wire carrying 10 A is placed inside a solenoid perpendicular to its axis where B = 0.27 T. Calculate the magnetic force on the wire and state its direction using the right-hand rule.

Two parallel wires A and B are separated by 4 cm and carry currents 8 A and 5 A respectively in the same direction. Estimate the force on a 10 cm section of wire A and state whether the wires attract or repel.

A horizontal wire carries 1 A in the east-to-west direction in a region where the horizontal component of the earth's field is 3 × 10⁻⁵ T directed from south to north. Find the force per unit length and its direction.

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