$A$ and $B$ are two conductors carrying a current $i$ in the same direction. $x$ and $y$ are two electron beams moving in the same direction
AThere will be repulsion between $A$ and $B$ attraction between $x$ and $y$
BThere will be attraction between $A$ and $B$, repulsion between $x$ and $y$
CThere will be repulsion between $A$ and $B$ and also $x$ and $y$
DThere will be attraction between $A$ and $B$ and also $x$ and $y$
Easy
Download our app for free and get started
BThere will be attraction between $A$ and $B$, repulsion between $x$ and $y$
b The force between two parallel current carrying conductors is given as,
$F=\frac{\mu}{4 \pi} \times \frac{2 I _1 L _2 \times 1}{a}$
Since, the current carrying conductors will attract each other, while electron beams will repel each other.Thus, the force is attraction between $A$ and $B$, is repulsion between $x$ and $y$.
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 proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbitals in a plane due to magnetic field perpendicular to the plane. Let $r_p, r_e$ and $r_{He}$ be their respective radii, then
The figure shows three situations when an electron moves with velocity $\vec v$ travels through a uniform magnetic field $\vec B$. In each case, what is the direction of magnetic force on the electron
A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at $\frac{a}{2}$ and $2$a from axis of the wire is:
A wire of length $L$ is bent in the form of a circular coil and current $i$ is passed through it. If this coil is placed in a magnetic field then the torque acting on the coil will be maximum when the number of turns is
A deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_{d}$ and $r_{\alpha}$ be their respective radii of circular path. The value of $\frac{r_{d}}{r_{\alpha}}$ is equal to
A conducting wire bent in the form of a parabola $y^2 = 2x$ carries a current $i = 2 A$ as shown in figure. This wire is placed in a uniform magnetic field $\vec B = - 4\,\hat k$ $Tesla$. The magnetic force on the wire is (in newton)
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
Electron of mass $m$ and charge $q$ is travelling with a speed along a circular path of radius $r$ at right angles to a uniform magnetic field of intensity $B$. If the speed of the electron is doubled and the magnetic field is halved the resulting path would have a radius
A square frame of side I carries a current $i$. The magnetic field at its centre is $B$. The same current is passed through a circular coil having the same perimeter as the square. The field at the centre of the circular coil is $B^{\prime}$. The ratio of $\frac{B}{B^{\prime}}$ is