Response:
(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v
Clarification:
Solution
Given that:
A refrigerator magnet with a depth of approximately 2 mm
The estimated magnetic field strength of the magnet is = 5 m T
The Area = 8 cm²
Now,
(A) The magnetic flux ΦB = BA
Therefore,
ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²
Thus,
ΦB = 4* 6 ^ ⁻6 T m²
(B) By employing Faraday's Law, the subsequent equation applies:
Ε = Bℓυ
Where,
ℓ = 2 cm equals 2 * 10 ^⁻2 m
B = 5 m T = 5 * 10 ^ ⁻3 T
υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s
Therefore,
Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v
E =2 * 10 ^ ⁻6 v
Answer:
Maximum emf = 5.32 V
Explanation:
Provided data includes:
Number of turns, N = 10
Radius of loop, r = 3 cm = 0.03 m
It made 60 revolutions each second
Magnetic field, B = 0.5 T
We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

For the maximum emf, 
Therefore,

Hence, the maximum emf generated in the loop is 5.32 V.
Response:
Clarification:
Provided
weight of disk 
diameter of disc 
weight of ring 
Force 



Overall moment of inertia
=Disc's moment of inertia +Ring's Moment of Inertia


At this point, Torque is 


Utilizing 
in this scenario


Answer:
Induced EMF is 2 x 10⁻³ volts
Explanation:
B = strength of the magnetic field aligning with the loop's axis = 1 T
= area change rate of the loop = 20 cm²/s = 20 x 10⁻⁴ m²
θ = the angle formed by the magnetic field and area vector = 0
E = the induced EMF across the loop
EMF can be calculated using the formula
E = B
E = (1) (20 x 10⁻⁴ )
E = 2 x 10⁻³ volts
E = 2 mV
Given that, the starting speed of the cells is 0 since they were at rest. The cell's acceleration is specified, along with time t = 700 ns. We aim to calculate the peak speed achieved by the cells and the distance covered during the acceleration. Let v signify the final velocity. Let d represent the distance traversed. We'll apply the equations of motion to find the solution.