Answer:
The induced current is 26.7 mA
Explanation:
Given;
area of the loop, A = 0.078 m²
initial magnetic field, B₁ = 3.8 T
change in the magnetic field strength, dB/dt = 0.24 T/s
The induced emf is calculated as;

The resistance of the loop = 0.7 Ω
The induced current is calculated as;

Answer:
a) 46.5º b) 64.4º
Explanation:
To solve this problem we will use the laws of geometric optics
a) For this part we will use the law of reflection that states that the reflected and incident angle are equal
θ = 43.5º
This angle measured from the surface is
θ_r = 90 -43.5
θ_s = 46.5º
b) In this part the law of refraction must be used
n₁ sin θ₁ = n₂. Sin θ₂
sin θ₂ = n₁ / n₂ sin θ₁
The index of air refraction is n₁ = 1
The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface
θ_s = 90 - θ
θ_s = 90- 43.5
θ_s = 46.5º
sin θ₂ = 1 / 1.68 sin 46.5
sin θ₂ = 0.4318
θ₂ = 25.6º
The angle with respect to the surface is
θ₂_s = 90 - 25.6
θ₂_s = 64.4º
measured in the fourth quadrant
Answer: 
Explanation:
Given
Charge discharged 
time taken 
Current is given as rate of change of discharge i.e.

Therefore, the average current is 