#wire

Resistance, Resistivity and Ohm's Law

A wire of length \( 2 \, \text{m} \) and resistance \( 5 \, \Omega \) is stretched to \( 4 \, \text{m} \). What is the n

**Resistance** R = ρ l/A, ρ resistivity (Ω·m), l length (m), A area (m²), ρ = m/(n e² τ) from Drude model, τ average collision time. Ohm's law V = I R holds when ρ constant, independent of V. Volume constant stretching l→2l implies A→A/2, so R' = ρ·2l/(A/2)=4R, resistance quadruples when length doubles at constant volume. Volume constant: l A = l' A' ⇒ A' = (A/2) . New resistance: R' = (rho l'/A') = (rho (2l)/(A/2)) = 4 (rho l/A) = 4R = 4 × 5 = 20 Ω . Applying I = n e A v_d, R = ρ l/A, R_t =

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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Resistance, Resistivity and Ohm's Law

A wire of length \( 2 \, \text{m} \) and resistance \( 4 \, \Omega \) is stretched to \( 4 \, \text{m} \). What is the n

**Resistance** R = ρ l/A, ρ resistivity (Ω·m), l length (m), A area (m²), ρ = m/(n e² τ) from Drude model, τ average collision time. Ohm's law V = I R holds when ρ constant, independent of V. Volume constant stretching l→2l implies A→A/2, so R' = ρ·2l/(A/2)=4R, resistance quadruples when length doubles at constant volume. Volume constant: l A = l' A' ⇒ A' = (A/2) . New resistance: R' = (rho l'/A') = (rho (2l)/(A/2)) = 4 (rho l/A) = 4R = 4 × 4 = 16 Ω . Applying I = n e A v_d, R = ρ l/A, R_t =

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 1.7 \, \text{m} \) carrying \( 7 \, \text{A} \) is at \( 60^\circ \) to a magnetic field of \( 0.2 \

**Force on current-carrying wire** in magnetic field is F = I l × B, magnitude F = I l B sinθ, I current (A), l length (m), B field (T), θ angle between current direction and B. Direction perpendicular to plane containing wire and B, given by Fleming's left-hand rule. F = I l B sin θ . F = 7 × 1.7 × 0.2 × sin 60° = 2.38 × 0.866 = 2.061 ≈ 2.06 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A wire of length \( 2 \, \text{m} \) carries a current of \( 5 \, \text{A} \) and is placed perpendicular to a magnetic

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. Force F = I l B sin θ , where θ = 90° , so sin θ = 1 . F = 5 × 2 × 0.4 = 4 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 2 \, \text{m} \) carrying \( 4 \, \text{A} \) is at \( 60^\circ \) to a magnetic field of \( 0.25 \,

**Force between parallel wires** per unit length is f = μ₀ I₁ I₂/(2π d), μ₀/2π = 2×10⁻⁷ T·m/A, d separation (m), attractive if currents same direction, repulsive if opposite. For I₁=5 A, I₂=7 A, d=0.04 m, f = 2×10⁻⁷×35/0.04 = 1.75×10⁻⁴ N/m, sign indicates repulsion for opposite directions. F = I l B sin θ . F = 4 × 2 × 0.25 × sin 60° = 2 × 0.866 = 1.732 ≈ 1.73 N . 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

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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PHYSICS

A copper wire carries 4.5 A with a drift speed of 1.8 × 10⁻⁴ m/s . If n = 8.5 × 10²⁸ m^{-3 and e = 1.6 × 10⁻

Given: A copper wire carries 4.5 A with a drift speed of 1.8 × 10⁻⁴ m/s . If n = 8.5 × 10²⁸ m^{-3 and e = 1.6 × 10⁻¹⁹ C, what is the cross-sectional area? These values define the system as per NCERT data. Formula: Drift speed: v_d = I/n e A. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Rearrange: A = I/n e v_d . Substitute: A = frac4.58.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.8 × 10⁻⁴. Calculate: A = 4.5/2.448 × 10⁵ approx 1.84 × 10⁻⁵ m² . Result: The computed value matches the expected outcome and confirms the

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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