A conducting rod of mass M with negligible electrical resistance slides on a pair of frictionless, horizontal, parallel, conducting rails separated by a distance L. The two rails are connected by an electrical resistance of R. A uniform magnetic field B is directed vertically upward in the entire region. a. Looking from above, is the conventional current flowing cw or cow? b. Determine the current through the resistor. c. Find the force necessary to keep the rod moving at a constant speed.

Explanation:

Given

mass of rod is M

Distance between rails is L

If the direction of magnetic field is directed upwards

If rod is moving right then,

Area enclosed is increasing therefore a Clockwise current will induce

and if rod is moving left then enclosed area is decreasing therefore an anticlockwise current will setup in the circuit

(b)Induced EMF will be

where v=velocity of rod

and current is

(c)Force necessary to move rod with constant velocity v is

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s