A \bar magnet has total length 2l = 20 units and the field point P is at a distance d = 10 units from the centre of the magnet. If the relative uncertainty of length measurement is 1\% , then uncertainty of the magnetic field at point P is:
Magnetic Field of a Bar Magnet diagram for Q10 - JEE Main 2025 Evening
A schematic mapping out the total \bar magnet layout alongside position coordinate node P.

Solution & Explanation

### Related Formula The standard expression for the magnetic field B on the axial path at distance r from the magnetic center is given by : B propto frac1r^3 Through logarithmic error differentiation: fracDelta BB = 3 cdot left(fracDelta rrright) ### Core Logic Depending on how the evaluation tracks parameter variables, two interpretations arise: **Method 1 (Approximating without considering variations in ell independently) [cite: 708, 709]:** If the spatial uncertainty is tied entirely to the radial component variable r : fracDelta BB = 3 times 1\% = 3\% **Method 2 (Accounting comprehensively for dimensional dependencies) [cite: 714, 715]:** If the error accumulation bounds combine tracking lengths alongside parameters: fracDelta BB = fracDelta ellell + 3left(fracDelta rrright) = 1\% + 3(1\%) = 4\% Both paths offer distinct structural insights depending on assumptions. The official valuation matrix accepts options reflecting both interpretations. ### Pattern Recognition In general engineering error evaluations, always look at power exponents. If an expression depends inversely on a cubed distance variable, fractional variation scales up by three \times the independent tracking variance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 12 Physics: Magnetism and Matter

Reference Study Guides

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Q55 jee_main_2024_29_jan_morning Magnetic Dipole
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. Answer: 6 to 6

Solution

### 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

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