The dimensional formula for magnetic flux is

Options

A. `[ML^(2)T^(-2)A^(-1)]`
B. `[ML^(3)T^(-2)A^(-2)]`
C. `[M^(0)L^(-2)T^(-2)A^(-2)]`
D. `[ML^(2)T^(-1)A^(2)]`

Answer

A

Explanation

Magnetic flux is given as
`phi=B S cos theta `……….i
Here is B magnetic field and S is area.
Magnetic force F on a charge q moving with velocity v is given as
`F=qvBsin beta`
So `B=F/(q v sin beta)` ……….ii
By equation I and ii
`phi=F/(qvsin beta)xxS cos theta`
Trigonometric functions are dimensitons so
`[phi]=[(FS)/(qv)]=(MLT^(-2)xxL^(2))/(ATxxLT^(-1))`
`=ML^(2)T^(-2)A^(-1)`