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Question: The dimension of magnetic field in M, L, T and C (Coulomb) is given as: A. \[ML{{T}^{-1}}{{C}^{-1}...

The dimension of magnetic field in M, L, T and C (Coulomb) is given as:
A. MLT−1C−1ML{{T}^{-1}}{{C}^{-1}}
B. MT2C−2M{{T}^{2}}{{C}^{-2}}
C. MT−1C−1M{{T}^{-1}}{{C}^{-1}}
D. MLT−2C−1ML{{T}^{-2}}{{C}^{-1}}

Explanation

Solution

Hint: The dimensional formula of magnetic field can be calculated using the formula B=FqvB=\dfrac{F}{qv} that has come from F=BqvF=Bqv. This is the force acting on a charge q moving with velocity v in a magnetic field B.

Complete step by step answer:
To calculate the dimensional formula of magnetic field we take the formula
B=FqvB=\dfrac{F}{qv}
This is the magnetic field B applied on a charge q, moving with a velocity v under the influence of a Magnetic Lorentz force F.
We must write each physical quantity used in its dimensional form:
Force F:
F=ma=[MLT−2]F=ma=[ML{{T}^{-2}}]
Charge q must be taken as Coulomb:
q=[C]q=[C]
Velocity v:
v=[LT−1]v=[L{{T}^{-1}}]
Substituting these in the first equation we get:
B=[MLT−2][C][LT−1]B=\dfrac{[ML{{T}^{-2}}]}{[C][L{{T}^{-1}}]}
On solving we get:
B=[MT−1C−1]B=[M{{T}^{-1}}{{C}^{-1}}]
Hence, the correct answer is option C. MT−1C−1M{{T}^{-1}}{{C}^{-1}}
Additional Information:
The SI units and Dimensional formula of some important physical quantities to remember are:
Work, Energy of all kinds = J,[M1L2T−2]J,[{{M}^{1}}{{L}^{2}}{{T}^{-2}}]
Power =W,[M1L2T−3]W,[{{M}^{1}}{{L}^{2}}{{T}^{-3}}]
Planck’s Constant (h) = Js,[M1L2T−1]Js,[{{M}^{1}}{{L}^{2}}{{T}^{-1}}]
Angular displacement (θ\theta)=rad,[M0L0T0]rad,[{{M}^{0}}{{L}^{0}}{{T}^{0}}].
Angular velocity (ω\omega)=rads−1[M0L0T0]rad{{s}^{-1}}[{{M}^{0}}{{L}^{0}}{{T}^{0}}]
Force constant (forcedisplacement\dfrac{\text{force}}{\text{displacement}}) = Nm−1,[M1L0T−2]N{{m}^{-1}},\left[ {{M}^{1}}{{L}^{0}}{{T}^{-2}} \right]
Coefficient of elasticity (stressstrain\dfrac{\text{stress}}{\text{strain}}) = Nm−2,[M1L−1T−2]N{{m}^{-2}},\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]
Angular frequency (ω)=,rads−1[M0L0T−1](\omega )=,rad{{s}^{-1}}[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]
Angular momentum Iω=kgm2s−1[M1L2T−1]I\omega =kg{{m}^{2}}{{s}^{-1}}[{{M}^{1}}{{L}^{2}}{{T}^{-1}}]

Note: Another method for solving this question would be by using the formula for force on a current carrying conductor placed in a magnetic field F=BILF=BILand therefore, B=FILB=\dfrac{F}{IL} .While solving dimensional formula questions students must note that every physical quantity must be expressed in its absolute units only.