JEE MainPhysicsUnits and Dimensions
In a new system of units, the speed of light in vacuum ( c ), Planck's constant ( h ), the universal gravitational constant ( G ), and magnetic field ( B ) are chosen as fundamental quantities. The dimensional formula of electric potential ( V ) in this new system is:
Options
- A[ c^ - 1 2 h^ 1 2 G^ 1 2 B ]
- B[ c^ - 3 2 h^ 1 2 G^ 1 2 B ]
- C[ c^ 11 2 h^ 1 2 G^ - 3 2 B⁻¹ ]
- D[ c^ 1 2 h^ - 1 2 G^ - 1 2 B ]
Correct answer
A. [ c^ - 1 2 h^ 1 2 G^ 1 2 B ]
Step-by-step solution
The dimensional formula for electric potential V is: [V] = [ Work ] [ Charge ] = M L^2 T⁻² A T = M L^2 T⁻³ A⁻¹ We first express the dimension of current ( A ) in terms of magnetic field ( B ). From the magnetic force equation F = qvB : [B] = M L T⁻² (A T)(L T⁻¹) = M T⁻² A⁻¹ A = M T⁻² B⁻¹ Substituting A into the dimensional formula for V : [V] = M L^2 T⁻³ (M T⁻² B⁻¹)⁻¹ = M L^2 T⁻³ M⁻¹ T^2 B = L^2 T⁻¹ B Now, we express L^2 T⁻¹ in terms of c , h , and G . Let: L^2 T⁻¹ = c^x h^y G^z Substituting the standard dimensions