The dimensional formula of frac{1}{2} mu_0 H^ | vedclass

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

Question

(C) The expression $\frac{1}{2} \mu_0 H^2$ represents the energy density of a magnetic field.Energy density is defined as energy per unit volume.The d

Vedclass · Accepted Answer

$[ML^{-1} T^{-2}]$

Vedclass · Answer

(C) The expression $\frac{1}{2} \mu_0 H^2$ represents the energy density of a magnetic field.Energy density is defined as energy per unit volume.The dimensional formula for energy is $[ML^2 T^{-2}]$ and for volume is $[L^3]$.Therefore, the dimensional formula for energy density is $\frac{[ML^2 T^{-2}]}{[L^3]} = [ML^{-1} T^{-2}]$.Alternatively, using the dimensions of the given quantities:$[\mu_0] = [MLT^{-2} A^{-2}]$ and $[H] = [AL^{-1}]$.Substituting these into the expression:$[\frac{1}{2} \mu_0 H^2] = [MLT^{-2} A^{-2}] 	imes [AL^{-1}]^2 = [MLT^{-2} A^{-2}] 	imes [A^2 L^{-2}] = [ML^{-1} T^{-2}]$.

The dimensional formula of $\frac{1}{2} \mu_0 H^2$ (where $\mu_0$ is the permeability of free space and $H$ is the magnetic field intensity) is:

Similar Questions

Two concentric loops $A$ and $B$ of same radius $R = 2 \pi \,cm = 2 \pi \times 10^{-2} \,m$ are placed at right angles to each other. If the currents flowing through $A$ and $B$ are $I_A = 3 \,A$ and $I_B = 4 \,A$ respectively, then the net magnetic field at their common centre is:

$A$ thin,straight conductor lies along the axis of a hollow conductor of radius $R$. The two carry equal currents in the same direction. The magnetic field $B$ is plotted against the distance $r$ from the axis. Which of the following best represents the resulting curve?

The unit of magnetic flux density (or magnetic induction) is:

Pick out the wrong statement.

Mark the correct statement.

Vedclass Test Series

Exam Paper Generator

Online Exam Module