Answer

Verified

459.3k+ views

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.

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Change the following sentences into negative and interrogative class 10 english CBSE