In the figure shown there are two semicircles of radii ${r_1}$ and ${r_2}$ in which a current $i$ is flowing. The magnetic induction at the centre $O$ will be
Medium
Download our app for free and get started
(c) The magnetic induction due to both semicircular parts will be in the same direction perpendicular to the paper inwards.
$\therefore B = {B_1} + {B_2} = \frac{{{\mu _0}i}}{{4{r_1}}} + \frac{{{\mu _0}i}}{{4{r_2}}} = \frac{{{\mu _0}i}}{4}\left( {\frac{{{r_1} + {r_2}}}{{{r_1}{r_2}}}} \right) \otimes $
Download our app
and 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.*
A galvanometer of resistance, $G,$ is shunted by a resistance $S$ $ohm$. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
A wire carrying current $I$ is bent in the shape $A\,B\,C\,D\,E\,F\,A$ as shown, where rectangle $A\,B\,C\,D\,A$ and $A\,D\,E\,F\,A$ are perpendicular to each other. If the sides of the rectangles are of lengths $a$ and $b,$ then the magnitude and direction of magnetic moment of the loop $A\,B\,C\,D\,E\,F\,A\,$ is
A straight section $PQ$ of a circuit lies along the $X$-axis from $x = - \frac{a}{2}$ to $x = \frac{a}{2}$ and carries a steady current $i$. The magnetic field due to the section $PQ$ at a point $X = + a$ will be
A proton of mass $1.67 \times {10^{ - 27}}\,kg$ and charge $1.6 \times {10^{ - 19}}\,C$ is projected with a speed of $2 \times {10^6}\,m/s$ at an angle of $60^\circ $ to the $X - $ axis. If a uniform magnetic field of $0.104$ $Tesla$ is applied along $Y - $ axis, the path of proton is
A proton and an $\alpha - $particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes $25$ $\mu \, sec$ to make $5$ revolutions, then the periodic time for the $\alpha - $ particle would be........$\mu \, sec$
A square-shaped conducting wire loop of dimension moving parallel to the $X$-axis approaches a square region of size $b(a < b)$, where a uniform magnetic field $B$ exists pointing into the plane of the paper (see figure). As the loop passes through this region, the plot correctly depicting its speed $v$ as a function of $x$ is
A magnet of magnetic moment $50\,\hat i\,A{\rm{ - }}{m^2}$ is placed along the $x-$ axis in a magnetic field $\overrightarrow B = (0.5\,\hat i + 3.0\hat j)\,T.$ The torque acting on the magnet is
A particle with charge $-Q$ and mass m enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ$. The direction of the motion of the $m$ particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{{QB}}$ . The time spent by the particle in the field will be