Ncert Exercises & Solutions

There are two current-carrying planar coil made each from identical wires of length L. $C_{1}$ is circular (radius R) and $C_{2}$ is 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.

Hint: Use the formula of frequency of oscillation of a bar magnet.

Step 1:

C_{1}

= circular coil of radius R, length L,

number of turns per unit length,

n_{1} = \frac{L}{2 πR}

C_{2}

= square of side a and perimeter L, number of turns per unit length,

n_{2} = \frac{L}{4 a}

\begin{array}{l} Magnetic moment of C_{1} \\ \Rightarrow m_{1} = n_{1} {IA}_{1} \\ Magnetic moment of C_{2} \end{array}

\Rightarrow m_{2} = n_{2} {IA}_{2}

m_{1} = \frac{L \cdot I \cdot {πR}^{2}}{2 πR}

m_{2} = \frac{L}{4 a} \cdot I \cdot a^{2}

m_{1} = \frac{LIR}{2} . . . (i)

m_{2} = \frac{LIa}{4} . . . (ii)

Step 2:

\begin{array}{l} Moment of inertia of C_{1} \Rightarrow I_{1} = \frac{{MR}^{2}}{2} . . . (iii) \\ Momont of inertia of C_{2} \to I_{2} = \frac{{Ma}^{2}}{12} . . . (iv) \\ Frequrncy of C_{1} \Rightarrow f_{1} = \frac{1}{2 π} \sqrt{\frac{m_{1} B}{I_{1}}} \end{array}

\begin{array}{l} Frequency of C_{2} \to f_{2} = \frac{1}{2 π} \sqrt{\frac{m_{2} B}{I_{2}}} \\ According to question, f_{1} = f_{2} \end{array}

\begin{array}{l} \frac{1}{2 π} \sqrt{\frac{m_{1} B}{I_{1}}} = \frac{1}{2 π} \sqrt{\frac{m_{2} B}{I_{2}}} \\ \frac{I_{1}}{m_{1}} = \frac{I_{2}}{m_{2}} or \frac{m_{2}}{m_{1}} = \frac{I_{2}}{I_{1}} \end{array}

Step 3:

Plugging the values by Eqs. (i), (ii), (iii) and (iv);

\begin{array}{l} \frac{LIa \cdot 2}{4 \times LIR} = \frac{{Ma}^{2} \cdot 2}{12 \cdot {MR}^{2}} \\ \frac{a}{2 R} = \frac{a^{2}}{6 R^{2}} \\ 3 R = a \end{array}

Thus, the value of a is 3R.

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