The dimensions [MLT^{-2}A^{-2}] belong to: | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The magnetic force on a current-carrying conductor is given by $F = BIl \sin 	heta$. Thus,the magnetic field $B = F / (Il)$.The dimensions of for

Vedclass · Accepted Answer

magnetic permeability

Vedclass · Answer

(B) The magnetic force on a current-carrying conductor is given by $F = BIl \sin 	heta$. Thus,the magnetic field $B = F / (Il)$.The dimensions of force $F$ are $[MLT^{-2}]$.The dimensions of current $I$ are $[A]$.The dimensions of length $l$ are $[L]$.Therefore,the dimensions of $B$ are $[MLT^{-2}] / ([A][L]) = [MT^{-2}A^{-1}]$.The magnetic permeability $\mu$ is related to $B$ and $H$ by $B = \mu H$,where $H$ has dimensions $[IL^{-1}] = [AL^{-1}]$.Thus,$\mu = B / H = [MT^{-2}A^{-1}] / [AL^{-1}] = [MLT^{-2}A^{-2}]$.Hence,the dimensions $[MLT^{-2}A^{-2}]$ correspond to magnetic permeability.

The dimensions $[MLT^{-2}A^{-2}]$ belong to:

Similar Questions

Obtain the relation between magnetisation $\vec{M}$ and magnetic intensity $\vec{H}$ for a solenoid.

When a piece of a ferromagnetic substance is placed in a uniform magnetic field,the flux density inside it is four times the flux density away from the piece. The magnetic permeability of the material is

Magnetization of the sample is defined as:

Susceptibility of $Mg$ at $300\, K$ is $1.2 \times 10^{-5}$. The temperature at which susceptibility will be $1.8 \times 10^{-5}$ is......$K$.

$A$ magnetic intensity of $500 \,A/m$ produces a magnetic flux of $2.4 \times 10^{-5} \,Wb$ in an iron bar of cross-sectional area $0.4 \,cm^2$. The magnetic permeability of the iron bar is

Vedclass Test Series

Exam Paper Generator

Online Exam Module