A small block of mass m, having charge q is placed on frictionless inclined plane making an angle θ with the horizontal. There exists a uniform magnetic field B parallel to the inclined plane but perpendicular to the length of spring. If m is slightly pulled on the inclined in downward direction and released, the time period of oscillation will be (assume that the block does not leave contact with the plane)

Question

A small block of mass m, having charge q is placed on frictionless inclined plane making an angle θ with the horizontal. There exists a uniform magnetic field B parallel to the inclined plane but perpendicular to the length of spring. If m is slightly pulled on the inclined in downward direction and released, the time period of oscillation will be (assume that the block does not leave contact with the plane) (A) 2 π √( m / K ) (B) 2 π √(2 m/K) (C) 2 π √( qB / K ) (D) 2 π √(q B/2 K)

Tardigrade Admin · Accepted Answer

There will be no effect of magnetic force on time period because the magnetic force will be perpendicular to the inclined plane.

$2 π \frac{m}{K}$

$2 π \frac{2 m}{K}$

$2 π \frac{qB}{K}$

$2 π \frac{qB}{2 K}$

There will be no effect of magnetic force on time period because the magnetic force will be perpendicular to the inclined plane.

2πKm​​

2πK2m​​

2πKqB​​

2π2KqB​​

There will be no effect of magnetic force on time period because the magnetic force will be perpendicular to the inclined plane.

$2 π \frac{m}{K}$

$2 π \frac{2 m}{K}$

$2 π \frac{qB}{K}$

$2 π \frac{qB}{2 K}$