A conducting rod of length \(L\) slides freely down an inclined plane, as shown in the figure. The plane is inclined at an angle \(\theta\) from the horizontal. A uniform magnetic field of strength \(B\) acts in the positive \(y\) -direction. Determine the magnitude and the direction of the current that would have to be passed through the rod to hold it in position on the inclined plane.

When charged particles, like electrons in a current-carrying conductor, move through a magnetic field, they experience a force known as the Lorentz force. This force can be described by the equation \( F = q(E + v \times B) \) where \( F \) is the force exerted on the particle, \( q \) is the charge of the particle, \( E \) is the electric field, \( v \) is the velocity of the particle, and \( B \) is the magnetic field. However, for a current-carrying wire, we drop the electric field term and the expression simplifies to \( F = q(v \times B) \), or in more general terms for a wire carrying a current I, the force is \( F = I(L \times B) \).
The direction of the Lorentz force is determined by the right-hand rule, which states that if you point your thumb in the direction of the current and your fingers in the direction of the magnetic field, your palm will face the direction of the force. This force is crucial in many technological applications such as electric motors and generators, and it's the central principle in the exercise of a rod sliding on an inclined plane.

The magnetic field strength, denoted by \( B \), represents the magnitude of the magnetic field's force on moving charges or currents within its area of influence. It's typically measured in teslas (T), with Earth's magnetic field being a gentle \( 10^{-5} \) T, while superconducting magnets can produce fields over 10 T. Stronger magnetic fields exert more force on moving charges or currents, which changes the behaviour of charged particles in a material or a current in a wire. In our exercise, the value of \( B \) determines the magnitude of the magnetic force that counteracts the gravitational force on the inclined plane, and figuring out its intensity is essential to calculate the current needed to keep the conducting rod in position.

The gravitational force is a force that attracts any objects with mass towards each other. For an object on an inclined plane, like the conducting rod in our scenario, the gravitational force can be represented as \( F_g = m \times g \times \sin(\theta) \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (9.81 m/s² on Earth's surface), and \( \theta \) is the angle of the inclined plane from the horizontal. This force tries to pull the rod down the incline, but in our problem, we are interested in balancing this force with the magnetic force to keep the rod from sliding down.

The current direction is an important factor in determining the direction of the magnetic force experienced by a conductor in a magnetic field, due to the Lorentz force. The established convention is that current direction is the direction in which positive charges would move, even though in most conductors, it's actually electrons (which have a negative charge) that are moving. In our exercise, the correct direction of the current is essential to ensure that the magnetic force counters the gravitational pull. As per the right-hand rule, the current must create a magnetic force upslope to keep the rod stationary on the inclined plane. If the current were to flow in the wrong direction, the resulting magnetic force would either be too weak or could further accelerate the rod's descent.