AP EAMCET201922 Apr 2019Morning ShiftPhysicsMagnetic Effects of CurrentActual
A small block of mass (20 ~g ) and charge (4 mC ) is released on a long smooth inclined plane of inclination angle of (45^ ), A uniform horizontal magnetic field of (1 ~T ) is acting parallel to the surface, as shown in the figure. The time from the start when the block loses contact with the surface of the plane is
Options
- A(2 ~s )
- B(3 ~s )
- C(5 ~s )
- D(6 ~s )
Correct answer
C. (5 ~s )
Step-by-step solution
Given inclination of plane, Mass of block, (m=20 ~g =0.02 ~kg ) charge on the block, (q=4 mC =4 10⁻³ C ) Magnetic field, (B=1 ~T ) Magnetic force on the charge particles, (F=B q v ) Particle will leave the inclined plane, when (F=m g B q v=m g v= m g q B ) Time taken to reached at the velocity (v ) is given by ( aligned & v=0+g t [ u=0, a=g ] & t= v g = m g q B g = m q B & t= 0.02 45^ 4 10⁻³ 1 =5 ~s [ 45^ =1 ] aligned )