A current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $ BC$ (radius $= b$) and $DA $ (radius $= a$) of the loop are joined by two straight wires $AB $ and $CD$. A steady current $I$ is flowing in the loop. Angle made by $AB$ and $CD$ at the origin $O$ is $30^o $. Another straight thin wire with steady current $I_1$ flowing out of the plane of the paper is kept at the origin.
Due to the presence of the current $I_1$ at the origin
AThe forces on $AB$ and $ DC$ are zero
BThe forces on $AD$ and $ BC$ are zero
CThe magnitude of the net force on the loop is given by $\frac{{{I_1}I}}{{4\pi }}{\mu _0}\left[ {2(b - a) + \frac{\pi }{3}(a + b)} \right]$
DThe magnitude of the net force on the loop is given by $\frac{{{\mu _0}I{I_1}}}{{24ab}}(b - a)$
AIEEE 2009, Medium
Download our app for free and get started
BThe forces on $AD$ and $ BC$ are zero
b KEY CONCEPT: $\vec{F}=I(\vec{\ell} \times \vec{B})$
The force on $A D$ and $B C$ due to current $I_{1}$ is zero. This is because the directions of current element $I \overrightarrow{d \ell}$ and magnetic field $\vec{B}$ are parallel.
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 thin flexible wire of length $\mathrm{L}$ is connected to two adjacent fixed points and carries a current $\mathrm{I}$ in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is
A particle of charge per unit mass $\alpha$ is released from origin with a velocity $\bar{v}=v_0 \vec{i}$ in a uniform magnetic field $\bar{B}=-B_0 \hat{k}$. If the particle passes through $(0, y, 0)$ then $y$ is equal to
A metallic rod of mass per unit length $0.5\; kg\; m^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^o$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction $0.25\; T$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is.....$A$
Two moving coil meters $M_1$ and $M_2$ having the following particulars :-
$R_1 = 10\,\Omega , N_1 = 30, A_1 = 3.6\times10^{-3}\, m^2, B_1 = 0.25\, T$
$R_2 = 14\,\Omega , N_2 = 42, A_2 = 1.8\times10^{-3}\, m^2, B_2 = 0.50\, T$
(The spring constants are identical for the two meters). Determine the ratio of voltage sensitivity of $M_2$ and $M_1$
Two long straight wires, each carrying a current $I$ in opposite direction are seperated by a distance $R$ . The magnetic induction at a point midway between the wire is
If two protons are moving with speed $v=4.5 \times 10^{5} \,m / s$ parallel to each other then the ratio of electrostatic and magnetic force between them