A metal cube with sides of length a=1cm is moving at velocity v0→=1m/sj^ across a uniform magnetic field B0→=5Tk^. The cube is oriented so that four of its edges are parallel to its direction of motion (i. e., the normal vectors of two faces are parallel to the direction of motion).Find E, the magnitude of the induced electric field inside the cube. Express your answer numerically, in newtons per coulomb.

the magnitude of the electric field is 1.25 N/C

Explanation:

The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s

ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V

Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube

ε = ∫E.ds

ε = Eds since E is always parallel to the side of the cube

= E∫ds  ∫ds = 4L since we have 4 sides

= E(4L)

= 4EL

So,4EL = 0.05 V

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

So, the magnitude of the electric field is 1.25 N/C

and 1and 1

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s