$A, B$ and $C$ are parallel conductors of equal length carrying currents $I, I$ and $2I$ respectively. Distance between $A$ and $B$ is $x$. Distance between $B$ and $C$ is also $x$. ${F_1}$ is the force exerted by $B$ on $A$ and $F_2$ is the force exerted by $B$ on $A$ choose the correct answer
${F_1} = 2{F_2}$
${F_2} = 2{F_1}$
${F_1} = {F_2}$
${F_1} = - {F_2}$
In figure the cube is of $40\,\, cm$ edge. Four straight segment of wire $ab, bc, cd$ and $da$ form a closed loop that carries a current $I = 5\,A$. A uniform magnetic field $0.02\,\,T$ is in $+y\,-$ direction ratio of magnetic force on segement $ab$ and $bc$ is
The force exerted by a magnetic field on a wire having length $L$ and current $I$ is perpendicular to the wire and given as $\left| F \right| = IL\left| B \right|$ . An experimental plot shows $(\vec F)$ as function of $L$ . The plot is a straight line with a slope $S = \left( {10 \pm 1} \right) \times {10^{ - 5}}\ AT$. The current in the wire is $\left( {15 \pm 1} \right)\ mA$ . The percentage error in $B$ is
The horizontal component of the earth's magnetic field at a certain place is $3.0 \times 10^{-5}\; T$ and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of $1 \;A$. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is $(a)$ east to west; $(b)$ south to north?
Three long current carrying wires $P, Q$ and $R$ placed perpendicular to plane of the paper. Magnetic force per unit length on wire $'R'$ is
A semi circular arc of radius $r$ and a straight wire along the diameter, both are carrying same current $i.$ Find out magnetic force per unit length on the small element $P$, which is at the centre of curvature.