Answer:
a).β=0.53
T
a).β=0.40
T
Explanation:
The magnetic field at distance 'r' from the center of toroid is given by:

a).

b).
The distance is the radius add the cross section so:




I would say clay, because its a solid
Answer:
(A) 2.4 N-m
(B) 
(C) 315.426 rad/sec
(D) 1741.13 J
(E) 725.481 rad
Explanation:
We have given mass of the disk m = 4.9 kg
Radius r = 0.12 m, that is distance = 0.12 m
Force F = 20 N
(a) Torque is equal to product of force and distance
So torque
, here F is force and r is distance
So 
(B) Moment of inertia is equal to 
So 
Torque is equal to 
So angular acceleration 
(C) As the disk starts from rest
So initial angular speed 
Time t = 4.6 sec
From first equation of motion we know that 
So 
(D) Kinetic energy is equal to 
(E) From second equation of motion
