Question

In the figure shown a circular coil of mass $m$ is hanged by two strings. A uniform magnetic field $B$ is set up in the horizontal direction. Then $\frac{T_2}{T_1}$ is equal to $\frac{x}{10}$ (use $\pi BIR = \frac{mg}{4}$). Here, $I$ is the current in the coil and $R$ is the radius. Also, current as seen from top is in clockwise sense. The value of $x$ is _______.

Answer 1

14

Answer 2

Explanation:
• In equilibrium the total weight is supported so that
  T₁ + T₂ = mg.
• Since the two strings are attached at diametrically opposite points (separated by a chord of length 2R) the couple due to the vertical tensions is
  (T₂ – T₁)·R.
• The magnetic moment of the circular coil is μ = I·(πR²) and with the coil at an angle 45° (so that sin 45 = 1/√2) the magnetic torque is
  τₘ = I·πR²·B·(1/√2).
• Equating the couple due to the tensions to the magnetic torque gives
  (T₂ – T₁)·R = I·πR²·B/√2
  ⇒ T₂ – T₁ = I·πR·B/√2.
• The problem supplies the relation
  πBIR = mg/4,
so that
  T₂ – T₁ = (mg/4)/√2 = mg/(4√2).
• Now, with
  T₁ + T₂ = mg    (1)
  T₂ – T₁ = mg/(4√2)  (2)
one solves to get:
  T₂ = ½[mg + mg/(4√2)]
  T₁ = ½[mg – mg/(4√2)].
• Their ratio is
  T₂/T₁ = [1 + 1/(4√2)]/[1 – 1/(4√2)].
Numerically, 1/(4√2) ≈ 0.1768 so that
  T₂/T₁ ≈ (1 + 0.1768)/(1 – 0.1768) ≈ 1.43.
• Since the problem states T₂/T₁ = x/10, we have
  x/10 ≈ 1.43  ⇒  x ≈ 14.3,
which rounds to 14.