Answer:
Radius, r = 0.00523 meters
Explanation:
It is given that,
Magnetic field, 
Current in the toroid, I = 9.6 A
Number of turns, N = 6
We need to find the radius of the toroid. The magnetic field at the center of the toroid is given by :
r = 0.00523 m
or
So, the radius of the toroid is 0.00523 meters. Hence, this is the required solution.
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
A: Buoyant force is equal to the weight
