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
Diffcult
Download our app for free and get started
(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 \times \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \, \times \left( {\frac{2}{3}i} \right)}}{a} - \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,\left( {\frac{i}{3}} \right) \times 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 $
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1A current carrying small loop behaves like a small magnet. If $A$ be its area and $M$ its magnetic moment, the current in the loop will beView Solution
- 2A horizontal rod of mass $10\, gm$ and length $10\, cm$ is placed on a smooth plane inclined at an angle of $60^\circ $ with the horizontal, with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field of induction $B$ is applied vertically downwards. If the current through the rod is $1.73$ $ampere$, then the value of $B$ for which the rod remains stationary on the inclined plane is......$Tesla$View Solution
- 3A charge $q$ is spread uniformly over an insulated loop of radius $r$ . If it is rotated with an angular velocity $\omega $ with respect to normal axis then the magnetic moment of the loop isView Solution
- 4If a proton, deutron and $\alpha - $ particle on being accelerated by the same potential difference enters perpendicular to the magnetic field, then the ratio of their kinetic energies isView Solution
- 5The resistance of a galvanometer is $50\, ohms$ and the current required to give full scale deflection is $100\,\mu A$. In order to convert it into an ammeter, reading upto $10\,A$, it is necessary to put a resistance ofView Solution
- 6A square loop of edge length $2 \mathrm{~m}$ carrying current of $2 \mathrm{~A}$ is placed with its edges parallel to the $\mathrm{x}-\mathrm{y}$ axis. A magnetic field is passing through the $x-y$ plane and expressed as $\vec{B}=B_0(1+4 x) \hat{k}$, where $\mathrm{B}_0=5 \mathrm{~T}$. The net magnetic force experienced by the loop is. . . . . . . $\mathrm{N}$.View Solution
- 7A voltmeter has resistance of $2000\, ohms$ and it can measure upto $2\,V$. If we want to increase its range to $10\, V$, then the required resistance in series will be ........... $\Omega $View Solution
- 8A galvanometer $G$ deflects full scale when a potential difference of $0.50 $ $V$ is applied. The internal resistance of the galvanometer $r_g$ is $25$ $ohms$. An ammeter is constructed by incorporating the galvanometer and an additional resistance $R_S$. The ammeter deflects full scale when a measurement of $2.0$ $A$ is made. The resistance $R_S$ is closest to : ................. $\Omega$View Solution
- 9A thin ring of $10\, cm$ radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40\,\pi \,rad\,{s^{ - 1}}$ about its axis, perpendicular to its plane. If the magnetic field at its centre is $3.8 \times {10^{ - 9}}\,T$, then the charge carried by the ring is close to $\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,N/{A^2}} \right)$View Solution
- 10View SolutionA uniform electric field and a uniform magnetic field are produced, pointed in the same direction. An electron is projected with its velocity pointing in the same direction