Figure shows a square loop ABCD with edge len | vedclass

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

Question

(b) According to question resistance of wire $ADC$ is twice that of wire $ABC$. Hence current flows through $ADC$ is half that of $ABC$ i.e. $\frac{{{

Vedclass · Accepted Answer

$\frac{{\sqrt 2 \,{\mu _0}i}}{{3\pi \,a}} \otimes $

Vedclass · Answer

(b) According to question resistance of wire $ADC$ is twice that of wire $ABC$. Hence current flows through $ADC$ is half that of $ABC$ i.e. $\frac{{{i_2}}}{{{i_1}}} = \frac{1}{2}$. Also ${i_1} + {i_2} = i$ $==>$ ${i_1} = \frac{{2i}}{3}$ and ${i_2} = \frac{i}{3}$

Magnetic field at centre $O$ due to wire $AB$ and $BC$ (part $1$ and $2$) ${B_1} = {B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2{i_1}\sin {{45}^o}}}{{a/2}} \otimes $$ = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,{i_1}}}{a} \otimes $

and magnetic field at centre $O$ due to wires $AD$ and $DC$ (i.e. part $3$ and $4$) ${B_3} = {B_4} = \frac{{{\mu _0}}}{{4\pi }}\frac{{2\sqrt 2 \,{i_2}}}{a}\odot$
Also $i_1 = 2i_2. \,So \,(B_1 = B_2) > (B_3 = B_4)$

Hence net magnetic field at centre $O$
${B_{net}} = ({B_1} + {B_2}) - ({B_3} + {B_4})$
$ = 2 	imes \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \, 	imes \left( {\frac{2}{3}i} ight)}}{a} - \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,\left( {\frac{i}{3}} ight) 	imes 2}}{a}$
$ = \frac{{{\mu _0}}}{{4\pi }}.\frac{{4\sqrt 2 \,i}}{{3a}}(2 - 1)\, \otimes = \frac{{\sqrt 2 \,{\mu _0}i}}{{3\pi \,a}} \otimes $

Figure shows a square loop $ABCD$ with edge length $a$. The resistance of the wire $ABC$ is $r$ and that of $ADC$ is $2r$. The value of magnetic field at the centre of the loop assuming uniform wire is

Similar Questions

One metre length of wire carries a constant current. The wire is bent to form a circular loop. The magnetic field at the centre of this loop is $B$. The same is now bent to form a circular loop of smaller radius to have four turns in the loop. The magnetic field at the centre of this new loop is

Two similar coils are kept mutually perpendicular such that their centres coincide. At the centre, find the ratio of the magnetic field due to one coil and the resultant magnetic field by both coils, if the same current is flown

A Rowland ring of mean radius $15\; cm\;3500$ turns of wire wound on a ferromagnetic core of relative permeability $800.$ What is the magnetic field $B$ (in $T$) in the core for a magnetizing current of $1.2\; A?$

A closely wounded circular coil of radius $5\,cm$ produces a magnetic field of $37.68 \times 10^{-4}\,T$ at its center. The current through the coil is $......A$. [Given, number of turns in the coil is $100$ and $\pi=3.14]$

A current of $0.1\, A$ circulates around a coil of $100$ $turns$ and having a radius equal to $5\,cm$. The magnetic field set up at the centre of the coil is ($\mu_0 = 4\pi \times 10^{-7} weber/amp-metre$)