$A$ long straight wire carrying a current of $25 \,A$ rests on a table. Another wire $PQ$ of length $1 \,m$ and mass $2.5 \,g$ carries the same current but in the opposite direction. The wire $PQ$ is free to slide up and down. To what height $h$ will wire $PQ$ rise (in $\,mm$)? (Take $g = 9.8 \,m/s^2$ and $\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A$)

For the circuit shown in the figure,the direction and magnitude of the force on the segment $PQR$ is:

The force per unit length on a straight wire carrying a current of $8 \ A$ making an angle of $30^{\circ}$ with a uniform magnetic field of $0.15 \ T$ is: (in $N \ m^{-1}$)

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?

If a straight current-carrying wire of linear density $0.12 \ kg \ m^{-1}$ is suspended in mid-air by a uniform horizontal magnetic field of $0.5 \ T$ normal to the length of the wire,then the current through the wire is (Acceleration due to gravity $= 10 \ m \ s^{-2}$; Neglect earth's magnetic field) (in $A$)

Two parallel wires of length $9 \, m$ each are separated by a distance $0.15 \, m$. If they carry equal currents in the same direction and exert a total force of $30 \times 10^{-7} \, N$ on each other,then the value of current must be ........ $A$. (in $.5$)