Question

C.G.S unit of magnetic field intensity is:
A. Tesla
B. Gauss
C. Weber
D. None of these

Answer 1

Hint: Magnetic field intensity is the strength of the magnetic field produced by a magnet or a current carrying conductor. When a charged particle (q) moving with some velocity (v), enters a magnetic field with magnetic field intensity (B), a force is exerted on the particle given by $\overrightarrow{{{F}_{B}}}=q\left( \overrightarrow{v}\times \overrightarrow{B} \right)$.

Complete step-by-step answer:
When a current is flowing in a wire or conductor, it produces a magnetic field in the surrounding. When another charged particle moving with some velocity enters into this magnetic field, a force is exerted on the particle such that it is deflected and continues its motion in a circular path. A magnet also produces a magnetic field in its surrounding. The strength of the magnetic field is called magnetic field intensity (B). It is a vector quantity. The magnetic field intensity at a point due to a current carrying wire is given by $\overrightarrow{dB}=\dfrac{{{\mu }_{\circ }}i}{4\pi }\left( \dfrac{\overrightarrow{dl}\times \widehat{r}}{{{r}^{2}}} \right)$. Here $\overrightarrow{dB}$ is the magnetic field intensity produced due to a small element of length dl of the wire, ${{\mu }_{\circ }}$ is a constant called magnetic permeability of space, $i$ is the current following in the wire, $\overrightarrow{dl}$ is the length of a small element of the wire, r is the distance of the point from the small element of the wire and $\widehat{r}$ is the unit vector of $\overrightarrow{r}$.
When a charged particle (q) with velocity $\overrightarrow{v}$ enters into a magnetic field of magnetic field intensity $\overrightarrow{B}$, then the magnetic force exerted on it is given by $\overrightarrow{{{F}_{B}}}=q\left( \overrightarrow{v}\times \overrightarrow{B} \right)$.
Therefore, the magnitude of the force will be $F=qvB\sin \theta $, where $\theta $ the angle between the vectors $\overrightarrow{v}$ and $\overrightarrow{B}$. We can rewrite the above equation as $B=\dfrac{F}{qv\sin \theta }$.
Let us now calculate the C.G.S unit of B. The C.G.S units of F, q and v are dyne, franklin and cm/s respectively. Therefore, the unit of B = $\dfrac{\text{dyne}}{\text{frankiln}\text{.cm}{{\text{s}}^{\text{-1}}}}\text{=}\dfrac{\text{dyn}\text{.s}}{\text{Fr}\text{.cm}}$, which is called Gauss.
Therefore, the correct option is (b) Gauss.

Note: Students have to be careful while reading a question. Some seem to be easy but the options can sometimes confuse. Like in this case, the first two options – tesla and gauss, both are the units of magnetic field intensity. Tesla is the S.I. unit and gauss is the C.G.S unit. In a hurry, students may mark option A.