The figure shows an equilateral triangle $ABC$ of side length '$l$' carrying currents as shown,placed in a uniform magnetic field '$B$' perpendicular to the plane of the triangle. The magnitude of the net magnetic force on the triangle is:

Two straight parallel wires,both carrying $10 \ A$ in the same direction,attract each other with a force of $1 \times 10^{-3} \ N$. If both currents are doubled,the force of attraction will be:

$A$ current of $1 \ A$ is flowing along the positive $x$-axis through a straight wire of length $0.5 \ m$ placed in a region of a magnetic field given by $\vec{B} = (2\hat{i} + 4\hat{j}) \ T$. The magnitude and the direction of the force experienced by the wire are,respectively:

The force exerted by a magnetic field on a wire having length $L$ and current $I$ is perpendicular to the wire and given as $|F| = IL|B|$. An experimental plot shows $|F|$ as a function of $L$. The plot is a straight line with a slope $S = (10 \pm 1) \times 10^{-5} \ T$. The current in the wire is $I = (15 \pm 1) \ mA$. The percentage error in $B$ is:

$A$ square loop of edge length $2 \ m$ carrying current of $2 \ A$ is placed with its edges parallel to the $x-y$ axis. $A$ magnetic field is passing through the $x-y$ plane and expressed as $\vec{B}=B_0(1+4x) \hat{k}$,where $B_0=5 \ T$. The net magnetic force experienced by the loop is . . . . . . $N$.

$A$ single turn current loop in the shape of a right-angled triangle with sides $5\,cm$,$12\,cm$,and $13\,cm$ carries a current of $2\,A$. The loop is placed in a uniform magnetic field of magnitude $0.75\,T$,whose direction is parallel to the current in the $13\,cm$ side of the loop. The magnitude of the magnetic force on the $5\,cm$ side is $\frac{x}{130}\,N$. The value of $x$ is $..........$