Answer:
83.3 Wb
Explanation:
The magnetic flux linkage through the coil is given by:
where
B is the magnetic field strength
A is the cross sectional area
N is the number of turns in the coil
is the angle between the direction of the field and the normal to the coil
In this problem:
B = 1.74 T
A = 0.133 m^2
N = 360
Therefore, the magnetic flux linkage is
Answer:
Acceleration = 2.35 m/
Speed = 8.67 m/s
Explanation:
The coefficient of friction , u =0.3
The angle of incline = 30°
The two forces acting on block are weight and friction.
weight along the incline = mg cos60° = = 0.5 mg
Friction along incline = umg cos30° = mg
Friction along incline = 0.26 mg
Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg
Acceleration = = 0.24 g = 2.35 m/
The height of incline = 8 m
Length of the inclined edge = 16 m
v= 8.67 m/s