A wire carrying current $I$ and other carrying $2I$ in the same direction produces a magnetic field $B$ at the mid point. What will be the field when $2I$ wire is switched off
Medium
Download our app for free and get started
(c) When two parallel conductors carrying current $I$ and $2I $ in same direction, then magnetic field at the midpoint is
$B = \frac{{{\mu _0}2l}}{{2\pi r}} - \frac{{{\mu _0}I}}{{2\pi r}} = \frac{{{\mu _0}I}}{{2\pi r}}$
When current $2I$ is switched off then magnetic field due to conductor carrying current $I$ is $B = \frac{{{\mu _0}I}}{{2\pi r}}$.
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.*
Consider the diagram shown below. A voltmeter of resistance $150\,\Omega$ is connected across $A$ and $B$. The potential drop across $B$ and $C$ measured by voltmeter is $...........\,V$
A singly ionized magnesium atom $(A=24)$ ion is accelerated to kinetic energy $5\,keV$ and is projected perpendicularly into a magnetic field $B$ of the magnitude $0.5\,T$. The radius of path formed will be___________ $cm$
$STATEMENT-1$ A vertical iron rod has a coil of wire wound over it at the bottom end. An alternating current flows in the coil. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil. Because
$STATEMENT- 2$ In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the upward direction.
The magnetic moment of a bar magnet is $0.5 \mathrm{Am}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \mathrm{~T}$. The work done in rotating it from its most stable to most unstable position is:
If the direction of the initial velocity of the charged particle is neither along nor perpendicular to that of the magnetic field, then the orbit will be
An electron is projected normally from the surface of a sphere with speed $v_0$ in a uniform magnetic field perpendicular to the plane of the paper such that its strikes symmetrically opposite on the sphere with respect to the $x-$ axis. Radius of the sphere is $'a'$ and the distance of its centre from the wall is $'b'$ . What should be magnetic field such that the charge particle just escapes the wall
A cylindrical conductor of radius $'R'$ carries a current $'i'$. The value of magnetic field at a point which is $R/4$ distance inside from the surface is $10\, T$. Find the value of magnetic field at point which is $4\,R$ distance outside from the surface