Two long straight wires $P$ and $Q$ carrying equal current $10\,A$ each were kept parallel to each other at $5\,cm$ distance. Magnitude of magnetic force experienced by $10\,cm$ length of wire $P$ is $F_1$. If distance between wires is halved and currents on them are doubled,force $F_2$ on $10\,cm$ length of wire $P$ will be :

$A$ wire carrying current '$I$' along the $x$-axis has length '$L$' and is kept in a magnetic field $\vec{B} = B(\hat{i} + 2\hat{j} - 2\hat{k}) \text{ T}$. The magnitude of the magnetic force acting on the wire is:

The horizontal component of the Earth's magnetic field at a certain place is $3 \times 10^{-5} \, T$ and the direction of the field is from geographic south to geographic north. $A$ very long straight conductor is carrying a steady current of $1 \, A$. Calculate the force per unit length on it if the direction of the current is from east to west (in $N/m$).

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?

$A$ wire carrying current $I$ along the $x$-axis has length $\ell$ and is kept in a magnetic field $\vec{B} = (\hat{i} + 2\hat{j} - 3\hat{k}) B \text{ Wb/m}^2$. The magnitude of the magnetic force acting on the wire is:

$A$ wire carrying a current $i$ is placed in a uniform magnetic field in the form of the curve $y = a \sin \left( \frac{\pi x}{L} \right)$,$0 \leq x \leq 2L$. The force acting on the wire is