Definition of Ampere

Answer: B. The defining separation is exactly one metre, in vacuum. This value is what makes the denominator 2π × 1 in the force-per-length expression of NCERT Class 12 Physics Chapter 4, page 124.

Why A is wrong: A is wrong because the definition fixes the separation at 1 m, not 1 cm; substituting d = 0.01 m would give a force per unit length 100 times larger than the defining value.

Why C is wrong: C is wrong because 1 mm is not the defining separation; with d = 10⁻³ m the force per unit length would be 1000 times the defining 2 × 10⁻⁷ N/m.

Why D is wrong: D is wrong because 10 cm is not the defining separation; the definition uses d = 1 m so that the arithmetic reduces cleanly to 2 × 10⁻⁷ N/m.

Answer: A. Substituting I₁ = I₂ = 1 A and d = 1 m into F/L = μ₀I₁I₂/(2πd) with μ₀ = 4π × 10⁻⁷ T·m/A gives exactly 2 × 10⁻⁷ N/m, as set out in NCERT Class 12 Physics Chapter 4, page 124.

Why B is wrong: B is wrong because 4π × 10⁻⁷ is the value of μ₀ in T·m/A, not the defining force per unit length; the 4π cancels against the 2π in the denominator during the substitution.

Why C is wrong: C is wrong because it drops the factor of 2: μ₀/(2π) = 2 × 10⁻⁷, not 1 × 10⁻⁷. The 4π/2π ratio equals 2.

Why D is wrong: D is wrong because the exponent is −7, not −5; μ₀ carries 10⁻⁷ and the 4π/2π ratio contributes only a factor of 2.

Answer: C. The defining conditions are infinitely long, straight, parallel conductors of negligible circular cross-section held in vacuum, because F/L = μ₀I₁I₂/(2πd) is derived for exactly that idealised geometry (NCERT Class 12 Physics Chapter 4, page 124).

Why A is wrong: A is wrong because finite length and a specified non-negligible cross-section both break the idealisation the force-per-length expression assumes; the cross-section must be negligible.

Why B is wrong: B is wrong because coiling the conductors into loops changes the geometry entirely, so the straight-parallel force-per-length expression no longer describes the arrangement.

Why D is wrong: D is wrong because the definition specifies vacuum, where μ₀ is the free-space permeability; a medium of relative permeability 2 would alter the force and hence the defining number.

Answer: D. F/L = μ₀I₁I₂/(2πd) = 2 × 10⁻⁷ × (3.00 × 3.00)/1.00 = 1.80 × 10⁻⁶ N/m. The defining 2 × 10⁻⁷ N/m scales with the product of the two currents (NCERT Class 12 Physics Chapter 4, page 123).

Why A is wrong: A is wrong because it scales the defining value by 3 instead of by the product 3.00 × 3.00 = 9.00; the force depends on I₁I₂, not on a single current.

Why B is wrong: B is wrong because 2.00 × 10⁻⁷ N/m is the value for 1 A in each wire; here the currents are 3.00 A each, so the result must be nine times larger.

Why C is wrong: C is wrong because it takes 9.00 as the mantissa but keeps the exponent −7 instead of writing 9.00 × 10⁻⁷ = ... — in fact 2 × 10⁻⁷ × 9.00 = 1.80 × 10⁻⁶, so the mantissa 9.00 forgets the leading factor of 2.

Answer: B. With d = 1.00 m, F/L = 2 × 10⁻⁷ I². Setting 2 × 10⁻⁷ I² = 8.00 × 10⁻⁷ gives I² = 4.00, so I = 2.00 A. This is the definition run backwards (NCERT Class 12 Physics Chapter 4, page 124).

Why A is wrong: A is wrong because it takes I² = 16.0 rather than 4.00; dividing 8.00 × 10⁻⁷ by 2 × 10⁻⁷ gives 4.00 for I², not for I.

Why C is wrong: C is wrong because it reads the ratio 8.00 × 10⁻⁷ / 10⁻⁷ = 8.00 as the current directly, ignoring both the factor of 2 and the square root.

Why D is wrong: D is wrong because it takes the square root of 2.00 instead of 4.00; the division by 2 × 10⁻⁷ must be done before the root, not after.

Answer: C. The defining quantity is a force per unit length, 2 × 10⁻⁷ N/m, not a total force. Because the conductors are infinitely long, the total force between them is unbounded; only the force per metre is finite and specifiable (NCERT Class 12 Physics Chapter 4, page 124).

Why A is wrong: A is wrong because vacuum is exactly what the definition specifies — μ₀ in the force expression is the free-space permeability.

Why B is wrong: B is wrong because parallel is correct; the force-per-length expression F/L = μ₀I₁I₂/(2πd) is derived for parallel conductors.

Why D is wrong: D is wrong because 1 m is the correct defining separation; substituting 1 cm would change the force per unit length by a factor of 100.

Answer: A. μ₀/(2π) = (4π × 10⁻⁷)/(2π) = 2 × 10⁻⁷ exactly. The definition is arithmetically clean precisely because μ₀ was historically assigned the value 4π × 10⁻⁷ T·m/A (NCERT Class 12 Physics Chapter 4, page 114).

Why B is wrong: B is wrong because the logic runs the other way: μ₀ is assigned the value 4π × 10⁻⁷ T·m/A and the 2 × 10⁻⁷ N/m figure follows from it by substitution, not the reverse.

Why C is wrong: C is wrong because d = 1 m contributes a factor of exactly 1 to the denominator, not 2; the 2 comes from the 4π/2π ratio.

Why D is wrong: D is wrong because the force per unit length is a single mutual quantity between the pair — it is not a one-wire result that gets doubled for the second wire.

Answer: B. F/L = μ₀I₁I₂/(2πd) = 2 × 10⁻⁷ × (2.00 × 2.00)/1.00 = 8.00 × 10⁻⁷ N/m, one substitution into the force law with the new currents (NCERT Class 12 Physics Chapter 4, page 123).

Why A is wrong: A is wrong because it scales the defining value by 2.00, the factor for a single current, instead of by the product of both scaling factors, 2.00 × 2.00 = 4.00.

Why C is wrong: C is wrong because it scales the defining value by 8.00, as if one current had been increased by a further factor of 2 beyond doubling.

Why D is wrong: D is wrong because it is the original defining value at 1.00 A in each wire; it does not reflect that both currents were doubled.