A straight conductor carrying current $i$ splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ of the loop is
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.*
Find the magnetic field at point $P$ due to a straight line segment $AB$ of length $6\, cm$ carrying a current of $5\, A$. (See figure) $(\mu _0 = 4p\times10^{-7}\, N-A^{-2})$
A charge of $1\,C$ is moving in a magnetic field of $0.5\,Tesla$ with a velocity of $10\,m/sec$ Perpendicular to the field. Force experienced is.....$N$
A square loop of side $2\, a ,$ and carrying current I, is kept in $XZ$ plane with its centre at origin. A long wire carrying the same current $I$ is placed parallel to the $z-$axis and passing through the point $(0, b, 0),(b>>a)$. The magnitude of the torque on the loop about $z-$axis is given by:
Two parallel long wires carry currents $i_1$ and $i_2$ with ${i_1} > {i_2}$. When the currents are in the same direction, the magnetic field midway between the wires is $10\, \mu T$. When the direction of $i_2$ is reversed, it becomes $40 \,\mu T$. the ratio ${i_1}/{i_2}$ is
An electron is projected with velocity $v_0$ in a uniform electric field $E$ perpendicular to the field. Again it is projetced with velocity $v_0$ perpendicular to a uniform magnetic field $B/$ If $r_1$ is initial radius of curvature just after entering in the electric field and $r_2$ is initial radius of curvature just after entering in magnetic field then the ratio $r_1:r_2$ is equal to