The radius of each coil of a Helmholtz galvanometer is 0.1 m and the number of turns in each is 25. When a current is passed through it, the deflection of the magnetic needle is observed as 45°. If the horizontal component of earth’s magnetic field is 0.314 × 10⁻⁴ T, then the value of current will be:
(A) 0.14 A
(B) 0.28 A
(C) 0.42 A
(D) 0.07 A

Given Data:
• Radius of each coil: r = 0.1 m
• Number of turns in each coil: N = 25
• Deflection of the magnetic needle: θ = 45°
• Horizontal component of Earth's magnetic field: B₀ = 0.314 × 10⁻⁴ T
• Helmholtz coil formula for magnetic field:
B=8μ0NI125rB = \frac{{8\mu_0 N I}}{{\sqrt{125} r}}
where μ₀ = 4π × 10⁻⁷ T·m/A (permeability of free space).
Step 1: Relation Between Fields and Tangent Law
The Tangent Law states that:
tan⁡θ=BB0\tan \theta = \frac{B}{B_0}
Substituting B=8μ0NI125rB = \frac{8\mu_0 N I}{\sqrt{125} r}:
tan⁡45∘=8μ0NI125rB0\tan 45^\circ = \frac{\frac{8\mu_0 N I}{\sqrt{125} r}}{B_0}
Since tan 45° = 1, we get:
8μ0NI125r=B0\frac{8\mu_0 N I}{\sqrt{125} r} = B_0
Solving for II:
I=B0125r8μ0NI = \frac{B_0 \sqrt{125} r}{8\mu_0 N}
Step 2: Substitute Given Values
I=(0.314×10−4)×125×(0.1)8×(4π×10−7)×25I = \frac{(0.314 \times 10^{-4}) \times \sqrt{125} \times (0.1)}{8 \times (4\pi \times 10^{-7}) \times 25}
Approximating 125≈11.18\sqrt{125} \approx 11.18:
I=(0.314×10−4)×(11.18)×(0.1)8×(4π×10−7)×25I = \frac{(0.314 \times 10^{-4}) \times (11.18) \times (0.1)}{8 \times (4\pi \times 10^{-7}) \times 25} I=(0.314×10−4×1.118)(8×4π×10−7×25)I = \frac{(0.314 \times 10^{-4} \times 1.118)}{(8 \times 4\pi \times 10^{-7} \times 25)} I=3.512×10−58×4π×10−7×25I = \frac{3.512 \times 10^{-5}}{8 \times 4\pi \times 10^{-7} \times 25} I=3.512×10−58×100π×10−7I = \frac{3.512 \times 10^{-5}}{8 \times 100\pi \times 10^{-7}} I=3.512×10−58×3.14×10−5I = \frac{3.512 \times 10^{-5}}{8 \times 3.14 \times 10^{-5}} I=3.51225.12I = \frac{3.512}{25.12} I≈0.14AI \approx 0.14 A
0.14A (Option A)\boxed{0.14 A \text{ (Option A)}}