JEE PYQ: Units & Measurements - Question ID c432df107932 (JEE Main 2020)
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Step-by-step Explanation
In electromagnetism, the expression represents the energy density stored in a magnetic field in vacuum. The constant is the permeability of free space.
To find its dimensions, we recall:
- Force between two current-carrying wires gives the dimension of :
- Magnetic field is defined via the Lorentz force , so
Step 1: Dimensions of
From the force formula we isolate : Taking dimensions:
Step 2: Dimensions of
From we get Hence
Step 3: Dimensions of
Square the dimension of :
Step 4: Combine to find
Divide by : The factor of is dimensionless, so the final dimension is
Common Traps & Exam Tip:1. Confusing with : Students sometimes use the dimensions of permittivity instead of permeability . 2. Forgetting the square on : Omitting the exponent leads to an incorrect power of mass or time. 3. Sign errors in exponents: Careful bookkeeping of negative signs in time and length exponents is essential.
Tip: Always write each step dimensionally and cancel units systematically to avoid arithmetic slips.
Related Questions from Units & Measurements
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(Least count of Vernier calliper )