Q152 / 160 · AP EAMCET 2019 (23 Apr Shift 1) · Physics
The resultant magnetic moment of three magnetic dipoles, each of the magnetic moment \(M\) shown in the arrangement is
- A\(\sqrt{2} M\)
- B\((\sqrt{2}+1) M\)
- Correct answerC\((\sqrt{2}-1) M\)
- D\(M\)
Solution
Three magnetic moments of magnetic dipole, are shown in the figure below as,
Here, \(M_1=\sqrt{M_A^2+M_B^2+2 M_A M_B \cos 90^{\circ}}\) \(\begin{aligned} & \Rightarrow \quad M_1=\sqrt{M^2+M^2+0} \\ & \Rightarrow \quad M_1=M \sqrt{2} \end{aligned}\) Now, the resultant, magnetic moment, \(\begin{aligned} & M_{\text {resultant }}=M_1+M_C \\ & \left(\because M_1 \text { and } M_C\right. \text { are } \\ \Rightarrow \quad & M_{\text {resultant }}=M \sqrt{2}+M \\ \Rightarrow \quad & M_{\text {resultant }}=(1+\sqrt{2}) \mathrm{M} \end{aligned}\) \((\because M_1 \text { and } M_C \text { are in the same direction) }\) \(\therefore\) correct option is (b).
Here, \(M_1=\sqrt{M_A^2+M_B^2+2 M_A M_B \cos 90^{\circ}}\) \(\begin{aligned} & \Rightarrow \quad M_1=\sqrt{M^2+M^2+0} \\ & \Rightarrow \quad M_1=M \sqrt{2} \end{aligned}\) Now, the resultant, magnetic moment, \(\begin{aligned} & M_{\text {resultant }}=M_1+M_C \\ & \left(\because M_1 \text { and } M_C\right. \text { are } \\ \Rightarrow \quad & M_{\text {resultant }}=M \sqrt{2}+M \\ \Rightarrow \quad & M_{\text {resultant }}=(1+\sqrt{2}) \mathrm{M} \end{aligned}\) \((\because M_1 \text { and } M_C \text { are in the same direction) }\) \(\therefore\) correct option is (b).
Hence, the correct answer is option: 'C'.