Magnetic Dipoles and Galvanometers (JEE Advanced)

Question 1

A thin insulated wire of length L carries a steady current I. In arrangement 1 the whole wire is bent into a single-turn square loop. In arrangement 2 the same whole wire is wound into a flat, closely

Accepted Answer

C. Square: side L/4, so m₁ = I(L²)/16.
Two-turn coil: each turn has circumference L/2, so radius r = L/(4π) and m₂ = 2 Iπ r² = (IL²)/(8π).
(m₁)/(m₂) = (L²/16)/(L²/(8π)) = π/2.
General result: a wire of length L wound into n circular turns has m = (IL²)/(4π n), so adding turns of a fixed wire reduces m.
Difficulty: Hard

Question 2

Use right-handed axes x, y, z. A closed loop of thin wire runs along six edges of a cube of edge a that has one corner at the origin O and its edges along the positive axes. The wire goes, in order, O

Accepted Answer

A. For any closed loop (planar or not) in a uniform field, the net force is zero and the torque is τ⃗ = m⃗×B⃗ with m⃗ = IA⃗, where the vector area is A⃗ = 1/2∮r⃗× dl⃗. For a polygon with vertices r⃗₁,…,r⃗ₙ: A⃗ = 1/2Σr⃗ᵢ×r⃗ᵢ₊₁.
Non-zero terms: (a,0,0)×(a,a,0) = a²ẑ; (a,a,0)×(a,a,a) = a²(x̂−ŷ); (a,a,a)×(0,a,a) = a²(−ŷ+ẑ); (0,a,a)×(0,0,a) = a²x̂.
Sum = a²(2,−2,2), so A⃗ = a²(x̂−ŷ+ẑ). (Check: addin

Question 3

A thin non-conducting disc of radius R and mass M has its mass distributed uniformly over its area. A total charge Q is spread over the disc with surface charge density proportional to the distance r

Accepted Answer

B. Ring between r and r+dr: charge dq = kr·2π r dr, current dI = dq ω/(2π) = kω r²dr, moment dm = π r²dI = π kω r⁴dr.
m = (π kω R⁵)/5 and Q = ∫₀^R 2π kr²dr = (2π kR³)/3, so m = 3/10Qω R².
Angular momentum (uniform mass): L = 1/2MR²ω.
m/L = (3Qω R²/10)/(MR²ω/2) = (3Q)/(5M).
The ratio exceeds Q/(2M) because the charge sits farther out than the mass.
Difficulty: Very hard

Magnetic Dipoles and Galvanometers: questions with solutions

Get the full chapter for ₹99