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There are two current-carrying planar coils made each from identical wires of length L. $C_1$ It is circular (radius R) and $C_2$ square (side a). They are so constructed that they have the same frequency of oscillation when they are placed in the same uniform B and carry the same current. Find 'a' in terms of R.

Answers (1)

For $C_1$ ,

Radius=R

length=L

Number of turns per unit length

$n_1=\frac{L}{2\pi R}$

For $C_2$

side=a

perimeter=L

Number of turns per unit length

$n_2=\frac{L}{4a}$

Let Magnetic moment $C_1$ be $m_1=n_1iA_1$ where i is the current in the coil and

The magnetic moment of $C_{2}$ be $m_2=n_2iA_2$ where i is the current in the coil

$m_1=L.i.\pi .\frac{R^2}{2\pi R } ; m_2=\frac{L}{4a}.i.a^2$

$m_1=\frac{LiR}{2 } ; m_2=\frac{Lia}{4}$

Moment of inertia of $C_1=I_1=\frac{MR^2}{}2$

Moment of inertia of $C_2=I_2=\frac{Ma^2}{12}$ ….ii where M is the mass of the coil

Frequency of $C_1=f_1=2\pi \sqrt{\frac{I_1}{m_1B}}$ …iii

Frequency of $C_2=f_2=2\pi \sqrt{\frac{I_2}{m_2B}}$ ….iv

According to the problem, $f_{1}=f_{2}$

$2\pi \sqrt{\frac{I_1}{m_1B}} =2\pi \sqrt{\frac{I_2}{m_2B}}$

$\frac{I_1}{m_1}=\frac{I_2}{m_2} or \frac{m_2}{m_1}=\frac{I_2}{I_1}$

On substituting the values from eqn i, ii, and iv we get

$Lia.\frac{2}{4 \times LiR}=Ma^{2}.\frac{2}{12 MR^2}$

$\frac{ a}{2R}=\frac{a^2}{6R^2 }$

$3R=a$

Thus, the value of 'a' is 3R.

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infoexpert24

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