Magnetic field due to currentHard
Question
A block of mass m & charge q is released on a long smooth inclined plane magnetic filed B is constant, uniform, horizontal and parallel to surface as shown. Find the time from start when block loses contact with the surface


Options
A.
B.
C.
D.None
Solution
v = (g sinθ)t
N = mg cosθ - qvB = 0 ⇒ t = mcotθ/qB
N = mg cosθ - qvB = 0 ⇒ t = mcotθ/qB
Create a free account to view solution
View Solution FreeMore Magnetic field due to current Questions
A current I flows through a circular coil of radius r the intensity of field at its centre is...1A current flows through an infinitel long straight wire. The magnetic field produced at a point 1m. away from it is : -...A particle of charge +q and mass m moving under the influence of uniform electric field and uniform magnetic field follo...A uniform magnetic field of magnitude 1T exists in region y > 0 is along +z axis as shown. A particle of charge 1 C i...Two thick wires and two thin wires, all of the same materials and same length form a square in the three different ways ...