A magnetic dipole experiences a torque of 80sqrt3mathrm~N~m$80\sqrt{3}\mathrm{~N~m}$ when placed in uniform magnetic field in such a way that dipole moment makes angle of 60^circ$60^{\circ}$ with magnetic field. The potential energy of the dipole is :

### Related Formula The magnetic potential (V$V$) at an axial location at distance r$r$ from the center of a magnetic dipole is given by: V = fracmu_04pi fracMr^2$$V = \frac{\mu_0}{4\pi} \frac{M}{r^2}$$ where M$M$ represents the magnetic moment. ### Core Logic Given values: V = 1.5 times 10^-5 mathrm~T cdot m$$V = 1.5 \times 10^{-5} \mathrm{~T \cdot m}$$ r = 20 mathrm~cm = 0.2 mathrm~m$$r = 20 \mathrm{~cm} = 0.2 \mathrm{~m}$$ fracmu_04pi = 10^-7 mathrm~T cdot m cdot A^-1$$\frac{\mu_0}{4\pi} = 10^{-7} \mathrm{~T \cdot m \cdot A^{-1}}$$ ### Step 1: Set up the Formula Substituting values into the axial expression: 1.5 times 10^-5 = 10^-7 times fracM(0.2)^2$$1.5 \times 10^{-5} = 10^{-7} \times \frac{M}{(0.2)^2}$$ 1.5 times 10^-5 = 10^-7 times fracM0.04$$1.5 \times 10^{-5} = 10^{-7} \times \frac{M}{0.04}$$ ### Step 2: Isolate and Compute M $ ### Step 2: Isolate and Compute M $M = frac1.5 times 10^-5 times 0.0410^-7$M = \frac{1.5 \times 10^{-5} \times 0.04}{10^{-7}}$ $ $M = frac0.06 times 10^-510^-7 = 0.06 times 10^2 = 6 mathrm~A cdot m^2$M = \frac{0.06 \times 10^{-5}}{10^{-7}} = 0.06 \times 10^2 = 6 \mathrm{~A \cdot m^2}$ Therefore, the magnetic moment of the dipole is $ Therefore, the magnetic moment of the dipole is $6 \mathrm{~A \cdot m^2}. ### Pattern Recognition Axial potential fields scale inversely with the square of distance ($. ### Pattern Recognition Axial potential fields scale inversely with the square of distance ($V \propto \frac{1}{r^2}$), analogous to electrostatic dipole potentials. Ensure the distance is converted directly to meters before squaring. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Magnetism and Matter