The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 mathrm~cm from its center is 1.5 times 10^-5 mathrm~T cdot m. The magnetic moment of the dipole is ________ mathrmA cdot m^2. left(text Given: fracmu_04 pi = 10^-7 mathrm~T cdot m cdot A^-1right)$

Numerical Answer Type:
Enter a numerical value Answer: 6 to 6 +4 marks

Solution & Explanation

### Related Formula The magnetic potential (V) at an axial location at distance r from the center of a magnetic dipole is given by: V = fracmu_04pi fracMr^2 where M represents the magnetic moment. ### Core Logic Given values: V = 1.5 times 10^-5 mathrm~T cdot m r = 20 mathrm~cm = 0.2 mathrm~m fracmu_04pi = 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 fracM0.04 ### Step 2: Isolate and Compute M M = frac1.5 times 10^-5 times 0.0410^-7 M = frac0.06 times 10^-510^-7 = 0.06 times 10^2 = 6 mathrm~A cdot m^2 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 (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

Reference Study Guides

More Magnetism and Matter Questions — jee_main_2024_29_jan_morning

Practice all Magnetism and Matter previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)