{Understanding the Forces}
The electrostatic force on a charge \( q \) is given by: \[ \vec{F_1} = q \vec{E} \] The magnetic force on a moving charge is given by: \[ \vec{F_2} = q (\vec{v} \times \vec{B}) \] Thus, the correct answer is (C).
The straight wire AB carries a current \(I\). The ends of the wire subtend angles \(\theta_1\) and \(\theta_2\) at the point \(P\) as shown in the figure. The magnetic field at the point \(P\) is:
In the given circuit, \(E_1 = E_2 = E_3 = 2V\) and \(R_1 = R_2 = 4\Omega\), then the current flowing through the branch AB is:
Five charges +q, +5q, -2q, +3q and -4q are situated as shown in the figure. The electric flux due to this configuration through the surface S is:
Two wires of the same material (Young’s modulus \( Y \)) and same length \( L \) but radii \( R \) and \( 2R \) respectively, are joined end to end and a weight \( W \) is suspended from the combination. The elastic potential energy in the system is:
A decimolar solution potassium ferrocyanide is 50% dissociated at 300 K. The osmotic pressure of solution is (R = 8.314 J K$^{-1}$ mol$^{-1}$):