Question

There are two current carrying planar coils made each from identical wires of length L. C₁ is circular (radius R) and C₂ is square (side a). They are so constructed that they have same frequency of oscillation when they are placed in the same uniform B and carry the same current. Find a in terms of R.

Answer 1

As the frequencies (ω) for both coil are given same

ω₁ = ω₂

`(2pi)/T_1 = (2pi)/T`

So the time period of both the coil C₁ and C₂ are equal so T₁ = T₂

`2pi sqrt(I_1/(m_1B)) = 2pi sqrt(I_2/(m_2B))`

`sqrt(I_1/m_1) = sqrt(I_2/m_2)`

As B is same in both coils so squaring both side we get,

`I_1/I_2 = m_1/m_2`  ......(I)

I₁, I₂ are the moment of inertia of coils C₁ and C₂ placed in same magnetic field B.

`I_1 = (mR^2)/2`  ......(II)

`I_2 = (ma^2)/12`  ......(III)

As the length of wire is same and identical so masses m₁ = m₂ = m

For circular-shaped coil magnetic moment

m₁ = n₁IA₁  ......[∵ Current (i) in both are same, i.e. I₁ = I₂ = I]

`m_1 = L/(2piR) * I * piR^2`   ......[∵ L = 2πRn₁]

`m_1 = (LIR)/2`  .......(IV)

For square-shaped coil, magnetic moment

`m_2 = n_2IA_2` as current I₁ = I₂ = I  ...(Given) = `L/(4a) I.a^2`

`m_2 = (LIa)/4`

Substitute II, III, IV, V in I

⇒ R²² × 12a² = R² × 4a

⇒ Ra = 13

⇒ 3R = a