A long straight wire carries a current of $\pi \,amp.$ The magnetic field due to it will be $5 \times {10^{ - 5}}{\rm{ }}\,weber/{m^{\rm{2}}}$ at what distance from the wire $[{\mu _o} = $ permeability of air]
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To know the resistance $G$ of a galvanometer by half deflection method, a battery of $emf\, V_E$ and resistance $R$ is used to deflect the galvanometer by angle $\theta $. If a shunt of resistance $S$ is needed to get half deflection then $G, R$ and $S$ related by the equation
A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction, extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is
A coil in the shape of an equilateral triangle of side $10\, {cm}$ lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field $20\, {mT}$. The torque acting on the coil when a current of $0.2\, {A}$ is passed through it and its plane becomes parallel to the magnetic field will be $\sqrt{{x}} \times 10^{-5} \,{Nm}$. The value of ${x}$ is ..... .
A solenoid of $N$ turn, $'l'$ length and $'r'$ radius of cross - section. If current $i$ flow in solenoid then magnetic field at axial mid point will be (where $l\, \simeq \,\,r$ )
A long solenoid has $100\,turns/m$ and carries current $i.$ An electron moves with in the solenoid in a circle of radius $2·30\,cm$ perpendicular to the solenoid axis. The speed of the electron is $0·046\,c$ ($c =$ speed of light). Find the current $i$ in the solenoid (approximate).....$A$
A current of $1\,A$ is flowing on the sides of an equilateral triangle of side $4.5\times10^{-2}\,m$ . The magnetic field at the centre of the triangle will be
A long solenoid has a radius $a$ and number of turns per unit length is $n$. If it carries a current $i$, then the magnetic field on its axis is directly proportional to
The magnetic field at the centre of a circular current carrying-conductor of radius $r$ is $B_c$. The magnetic field on its axis at a distance $r$ from the centre is $B_a$. The value of $B_c : B_a$ will be :-