99 Percentile Qs Bank for JEE MainPhysicsUnits and Dimensions
The dimension of E^2 ₀ in mass (M) , length (L) and time (T) is ( E= electric field, ₀= permeability of free space)
Options
- A[ M ^2 ~L ^3 ~T ⁻² ~A ^2 ]
- B[ MLT ⁻⁴ ]
- C[ ML ^3 ~T ⁻² ]
- D[ ML ^4 ~T ⁻⁴ ]
Correct answer
B. [ MLT ⁻⁴ ]
Step-by-step solution
We know that, aligned & Electric field intensity = Electrostatic force Electric charge & E= F q aligned Permeability of Free Space ( ₀ ) Dimension Calculation Magnetic field intensity in at a point in space due to an infinitely long current carrying conducting wire Now, the magnetic force experienced by a current carrying conductor, when placed in a uniform external magnetic field is given by F=I L B ( ) B= F I L ( ) Put this value of B into Eq. (ii) to get aligned & ₀= 2 r [ F I L ( ) ] I = 2 r F I^2 L ( ) & [ ₀ ]