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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.
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Solution
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
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