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#inclined field

3 public questions tagged with this topic.

Magnetic Field Due to Current - Straight Wire and Circular Loop

A rectangular loop of area \( 0.04 \, \text{m}^2 \) with 18 turns carries \( 2.8 \, \text{A} \) in a field of \( 0.9 \,

**Magnetic field due to long straight wire** at distance r is B = μ₀ I/(2π r), μ₀ = 4π×10⁻⁷ T·m/A, direction circular around wire given by right-hand grip rule. For I = 18 A, r = 0.15 m, B = 2×10⁻⁷×18/0.15 = 2.4×10⁻⁵ T, illustrating 1/r dependence. tau = N I A B sin θ , where θ = 60° to plane means sin 30° with normal. tau = 18 × 2.8 × 0.04 × 0.9 × sin 60° = 1.8144 × 0.866 = 1.5712 ≈ 1.57 N m . Using F = q v B sinθ, F = I l B sinθ, B =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Field Due to Current - Straight Wire and Circular Loop

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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 0.5 \, \text{m} \) carrying \( 4 \, \text{A} \) is placed at \( 60^\circ \) to a magnetic field of \

**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. Force F = I l B sin θ . F = 4 × 0.5 × 0.3 × sin 60° = 2 × 0.3 × 0.866 = 0.5196 ≈ 0.52 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 > Force on Current-Carrying Conductor and Between Parallel Wires

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A rectangular loop of area \( 0.03 \, \text{m}^2 \) with 20 turns carries \( 2 \, \text{A} \) in a field of \( 0.7 \, \t

**Magnetic field due to long straight wire** at distance r is B = μ₀ I/(2π r), μ₀ = 4π×10⁻⁷ T·m/A, direction circular around wire given by right-hand grip rule. For I = 18 A, r = 0.15 m, B = 2×10⁻⁷×18/0.15 = 2.4×10⁻⁵ T, illustrating 1/r dependence. tau = N I A B sin θ , where θ = 45° to plane means sin 45° with normal. tau = 20 × 2 × 0.03 × 0.7 × sin 45° = 0.84 × 0.707 = 0.5939 ≈ 0.59 N m . Using F = q v B sinθ, F = I l B sinθ, B =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Field Due to Current - Straight Wire and Circular Loop

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