Solenoid, Toroid and Ampere's Law

Solenoid, Toroid and Ampere's Law

A circular coil of 70 turns and radius \( 7 \, \text{cm} \) carries a current of \( 0.8 \, \text{A} \). What is the magn

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 70 × 0.8/2 × 0.07) = (22.4 π × 10⁻⁶/0.14) = 1.6 π × 10⁻⁴ ≈ 5.03 × 10⁻⁴ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B

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Solenoid, Toroid and Ampere's Law

A wire of length \( 1.5 \, \text{m} \) carrying \( 8 \, \text{A} \) is at \( 60^\circ \) to a magnetic field of \( 0.5 \

**Solenoid field** inside long solenoid is B = μ₀ n I, n = N/L turns per meter (m⁻¹), uniform and parallel to axis, outside negligible for long solenoid because fields from opposite sides cancel. For n = 1200 m⁻¹, I = 1 A, B = 4π×10⁻⁷×1200 = 1.51×10⁻³ T = 1.51 mT. F = I l B sin θ . F = 8 × 1.5 × 0.5 × sin 60° = 12 × 0.5 × 0.866 = 5.196 ≈ 5.2 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R)

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Solenoid, Toroid and Ampere's Law

A circular loop of radius \( 0.13 \, \text{m} \) with 20 turns carries a current of \( 3.5 \, \text{A} \). What is the m

**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 20 × 3.5/2 × 0.13) = (28 π × 10⁻⁶/0.26) = 1.0769 π × 10⁻⁴ ≈ 3.38 × 10⁻⁴ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B =

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Solenoid, Toroid and Ampere's Law

Which factor does not affect the magnetic field produced by a current element according to the Biot-Savart law?

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. The Biot-Savart law states dB ∝ (I dl × r/r³) . The field depends on current, length of the element, distance, and angle, but not on the mass of the conductor. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ

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Solenoid, Toroid and Ampere's Law

A long wire carries \( 12 \, \text{A} \). At what distance is the magnetic field \( 2.4 \times 10^{-6} \, \text{T} \)? (

**Solenoid field** inside long solenoid is B = μ₀ n I, n = N/L turns per meter (m⁻¹), uniform and parallel to axis, outside negligible for long solenoid because fields from opposite sides cancel. For n = 1200 m⁻¹, I = 1 A, B = 4π×10⁻⁷×1200 = 1.51×10⁻³ T = 1.51 mT. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 12/2 π × 2.4 × 10⁻⁶) = (48 × 10⁻⁷/4.8 × 10⁻⁶) = 1 m . Using F = q v B sinθ, F = I l B sinθ, B =

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Solenoid, Toroid and Ampere's Law

A proton moves with a speed of \( 2 \times 10^6 \, \text{m/s} \) perpendicular to a uniform magnetic field of \( 0.5 \,

**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. Radius r = (mv/qB) . Substitute: r = (1.67 × 10⁻²⁷ × 2 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (3.34 × 10⁻²¹/8 × 10⁻²⁰) = 4.175 × 10⁻² m = 4.18 cm . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R)

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Solenoid, Toroid and Ampere's Law

A circular loop of radius \( 0.15 \, \text{m} \) with 15 turns carries a current of \( 2 \, \text{A} \). What is the mag

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 15 × 2/2 × 0.15) = (12 π × 10⁻⁶/0.3) = 4 π × 10⁻⁵ ≈ 1.26 × 10⁻⁴ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B

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Solenoid, Toroid and Ampere's Law

What happens to the torque on a rectangular current loop if the magnetic field direction is reversed?

**Solenoid field** inside long solenoid is B = μ₀ n I, n = N/L turns per meter (m⁻¹), uniform and parallel to axis, outside negligible for long solenoid because fields from opposite sides cancel. For n = 1200 m⁻¹, I = 1 A, B = 4π×10⁻⁷×1200 = 1.51×10⁻³ T = 1.51 mT. Torque is given by boldsymboltau = m × B . Reversing the magnetic field B reverses the direction of the torque vector, but its magnitude remains the same. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ

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Solenoid, Toroid and Ampere's Law

A straight wire of length \( 1.2 \, \text{m} \) carries a current of \( 6 \, \text{A} \) perpendicular to a uniform magn

**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. Force F = I l B sin θ , where θ = 90° , so sin θ = 1 . F = 6 × 1.2 × 0.25 = 1.8 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =

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Solenoid, Toroid and Ampere's Law

An electron moves at \( 5 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.2 \, \text{T} \). What

**Solenoid field** inside long solenoid is B = μ₀ n I, n = N/L turns per meter (m⁻¹), uniform and parallel to axis, outside negligible for long solenoid because fields from opposite sides cancel. For n = 1200 m⁻¹, I = 1 A, B = 4π×10⁻⁷×1200 = 1.51×10⁻³ T = 1.51 mT. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 5 × 10⁶ × 0.2 = 1.6 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π

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Solenoid, Toroid and Ampere's Law

A proton moves at \( 4 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.05 \, \text{T} \). What is the radi

**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 4 × 10⁷/1.6 × 10⁻¹⁹ × 0.05) = (6.68 × 10⁻²⁰/8 × 10⁻²¹) = 8.35 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I A B

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Solenoid, Toroid and Ampere's Law

A solenoid has 850 turns per meter and carries a current of \( 1.6 \, \text{A} \). What is the magnetic field inside it?

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. Magnetic field B = μ₀ n I . B = 4 π × 10⁻⁷ × 850 × 1.6 = 5.44 π × 10⁻⁴ ≈ 1.71 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

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