Question

There are two current carrying planar coils made each from identical wires of length L. C is circular (radius R) and C2 is square (side a). They are so constructed that they have same frequencies of oscillation when they are placed in the same uniform magnetic field B and carry the same current I. The relation between a and R is

• A. a = 2R
• B. a = 3R
• C. 3a = R
• D. a = 4R

Answer 1

Answer: (B)

a = 3R

Answer 2

: For circular coil,

and magnetic moment, $\mathrm { M } _ { 1 } = \mathrm { n } _ { 1 } \mathrm { IA } _ { 1 }$

For square coil

and magnetic moment,

Now moment of inertia about an axis through the diameter of circular coil and moment of inertia of square coil about an axis passing through its centere and parallel to edge of square

Then, frequency of oscillation of $\mathrm { C } _ { 1 } \cdot \omega _ { 1 } = \sqrt { \frac { \mathrm { M } _ { 1 } \mathrm {~B} } { \mathrm { I } _ { 1 } } }$ and frequency of oscillation of $\mathrm { C } _ { 2 } \cdot \omega _ { 2 } = \sqrt { \frac { \mathrm { M } _ { 2 } \mathrm {~B} } { \mathrm { I } _ { 2 } } }$

As, $\omega _ { 1 } ^ { 2 } = \omega _ { 2 } ^ { 2 } \Rightarrow \frac { M _ { 1 } } { I _ { 1 } } = \frac { M _ { 2 } } { I _ { 2 } }$

$= \frac { \frac { \text { LIR } } { 2 } } { \frac { \mathrm { MR } ^ { 2 } } { 2 } } = \frac { \frac { \mathrm { LIa } } { 4 } } { \frac { \mathrm { Ma } ^ { 2 } } { 12 } } \Rightarrow \mathrm { a } = 3 \mathrm { R }$