Question:

In the formula, X=3YZ2,X and Z have dimensions of capacitance and magnetic field respectively. What are the dimensions of Y in the MKSQ system?

Updated On: Oct 1, 2024
  • (A) [M3 L1 T3Q4]
  • (B) [M3 L2 T4Q4
  • (C) [M2L2 T4Q4]
  • (D) [M3 L2 T4Q]
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The Correct Option is B

Solution and Explanation

Explanation:
The magnetic force (F) experienced by a moving charged particle(q) in a uniform magnetic field (B) is given byF=qvBsinθB=Fqvsinθwhere, θ is angle between the magnetic field and the velocity of the particle.The dimensions of charge, force and velocity are[q]=[Q][F]=[MLT2][v]=[LT1]sinθ is a dimensionless quantity. Now, the dimension of magnetic field is[B]=[F]|q|vsinθ]=[MLT2][Q][LT1]=[MT1Q1]The energy stored in a capacitor is given byE=q22Cwhere, C is capacitance.The dimension of energy is[E]=[ML2T2]Thus, the dimension of capacitance is[C]=[q2][E][C]=[Q2][ML2T2][C]=[M1L2T2Q2]Given: X=3YZ2 where, X and Z have the dimensions of capacitance and magnetic field respectively. Thus, [X]=[C]=[M1L2T2Q2[z]=[B]=[MT1Q1]Therefore, the dimension of Y is [Y]=[X][Z]2[Y]=[M1 L2 T2Q2][MT1Q1]2[Y]=[M1L2T2Q2]|M2T2Q2|[Y]=[M3L2T4Q4]Hence, the correct option is (B).
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