InterviewSolution
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InterviewSolution
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There are two current carrying planar coils made each from identical wires of length L. `C_1` is the circular (radius R) and `C_2` is square (side a). They are so constructed that they have same frequency of oscillation when they are placed in the same uniform `vecB` and carry the same current i. Find a in terms of R. |
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Answer» For circular coil `C_1`, `n_1=(L)/(2piR)`, Magnetic moment, `M_1=n_1iA_1=(L)/(2piR)xxixxpiR^2=(LiR)/(2)` For square coil `C_2`, `n_2=(L)/(4a)`, Magnetic moment, `M_2=n_2iA_2=(L)/(4a)xxixxa^2=(Lia)/(4)` Moment of inertia of circular coil about the diameter as axis, `I_1=(massxx (radius)^2)/(2)=(mR^2)/(2)` Moment of inertia of square coil about an axis passing through its centre parallel to breadth `I_2=(ma^2)/(12)`. Time period of oscillation of the magnet in magnetic field is given by `T=2pisqrt(I)/(MB)` and `omega=(2pi)/(T)=sqrt((MB)/(I))` `:.` `omega_1^2=(M_1B)/(I_1)` and `omega_2^2=(M_2B)/(I_2)` Given, `omega_1^2=omega_2^2`, so `(M_1)/(I_1)=(M_2)/(I_2)` or `(LIR//2)/(mR^2//2)=(LIa//4)/(ma^2//12)` On solving, `a=3R` |
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