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Magnetic Effects of Current and Magnetism - Moving Charges and Magnetic Field (Biot-Savart & Ampere's Law)

Grade 12ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Oersted's Experiment: Demonstrated that a current-carrying conductor produces a magnetic field around it, with the direction determined by the Right Hand Thumb Rule.

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Biot-Savart Law: Relates the magnetic field dBd\mathbf{B} to the current element IdlI d\mathbf{l}. The field is proportional to the current and the sine of the angle between the element and the position vector, and inversely proportional to the square of the distance.

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Magnetic Field of a Circular Loop: At the center, the field is maximum and directed perpendicular to the plane of the loop. For NN turns, the field is multiplied by NN.

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Ampere's Circuital Law: The line integral of the magnetic field B\mathbf{B} around any closed path (Amperian loop) is equal to μ0\mu_0 times the total current II threading through the loop.

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Solenoid and Toroid: A solenoid produces a strong, uniform magnetic field inside along its axis. An ideal toroid confines the magnetic field within its core, with zero field outside or in the central hole.

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Permeability of Free Space: The constant μ0\mu_0 has a value of 4π×10−7 T⋅m/A4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}.

📐Formulae

dB=μ04πI(dl×r)r3d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I (d\mathbf{l} \times \mathbf{r})}{r^3}

dB=μ04πIdlsin⁡θr2dB = \frac{\mu_0}{4\pi} \frac{I dl \sin \theta}{r^2}

Bcenter=μ0I2RB_{center} = \frac{\mu_0 I}{2R}

Baxis=μ0IR22(R2+x2)3/2B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}

∮B⋅dl=μ0Ienclosed\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enclosed}

Bstraight wire=μ0I2πrB_{straight\,wire} = \frac{\mu_0 I}{2\pi r}

Bsolenoid=μ0nIwhere n=NLB_{solenoid} = \mu_0 n I \quad \text{where } n = \frac{N}{L}

Btoroid=μ0NI2πrB_{toroid} = \frac{\mu_0 N I}{2\pi r}

💡Examples

Problem 1:

Calculate the magnetic field at the center of a circular coil of 100100 turns and radius 10 cm10 \text{ cm} carrying a current of 0.1 A0.1 \text{ A}.

Solution:

B=μ0NI2RB = \frac{\mu_0 N I}{2R} B=4π×10−7×100×0.12×0.1B = \frac{4\pi \times 10^{-7} \times 100 \times 0.1}{2 \times 0.1} B=2π×10−5 T≈6.28×10−5 TB = 2\pi \times 10^{-5} \text{ T} \approx 6.28 \times 10^{-5} \text{ T}

Explanation:

We use the formula for the magnetic field at the center of a circular coil, accounting for NN turns. Convert the radius from cm\text{cm} to m\text{m} (0.1 m0.1 \text{ m}) before calculation.

Problem 2:

A long straight wire carries a current of 35 A35 \text{ A}. Find the magnitude of the magnetic field at a point 20 cm20 \text{ cm} from the wire.

Solution:

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r} B=4π×10−7×352π×0.2B = \frac{4\pi \times 10^{-7} \times 35}{2\pi \times 0.2} B=2×10−7×350.2=3.5×10−5 TB = \frac{2 \times 10^{-7} \times 35}{0.2} = 3.5 \times 10^{-5} \text{ T}

Explanation:

Applying Ampere's Law for an infinite straight wire. The distance rr must be in meters (0.2 m0.2 \text{ m}). The factor μ02π\frac{\mu_0}{2\pi} simplifies to 2×10−72 \times 10^{-7}.

Problem 3:

A solenoid of length 0.5 m0.5 \text{ m} has a radius of 1 cm1 \text{ cm} and is made up of 500500 turns. It carries a current of 5 A5 \text{ A}. What is the magnitude of the magnetic field inside the solenoid?

Solution:

n=NL=5000.5=1000 turns/mn = \frac{N}{L} = \frac{500}{0.5} = 1000 \text{ turns/m} B=μ0nI=4π×10−7×1000×5B = \mu_0 n I = 4\pi \times 10^{-7} \times 1000 \times 5 B=2π×10−3 T≈6.28×10−3 TB = 2\pi \times 10^{-3} \text{ T} \approx 6.28 \times 10^{-3} \text{ T}

Explanation:

First, calculate the number of turns per unit length (nn). Then use the formula for the interior field of a long solenoid. Note that the radius is irrelevant if the solenoid is long enough.