Answer:
The frequency of the coil is 7.07 Hz
Explanation:
Given;
number of turns of the coil, 200 turn
cross sectional area of the coil, A = 300 cm² = 0.03 m²
magnitude of the magnetic field, B = 30 mT = 0.03 T
Maximum value of the induced emf, E = 8 V
The maximum induced emf in the coil is given by;
E = NBAω
Where;
ω is angular frequency = 2πf
E = NBA(2πf)
f = E / 2πNBA
f = (8) / (2π x 200 x 0.03 x 0.03)
f = 7.07 Hz
Therefore, the frequency of the coil is 7.07 Hz
Answer:
The total number of significant figures is twelve.
Explanation:
Answer:
your answer is: electron → carbon atom → quantum dot → E. coli bacteria cell → comma
Explanation:
Calculate the magnitude of the linear momen- tum for each of the following cases a) a proton with mass 1.67 × 10-27 kg mov- ing with a velocity of 6 × 106 m/s. Answer in units of kg · m/s.
Answer:
The torque on the loop is 
Nm
Explanation:
Given:
Current 
A
Magnetic field 
T
Area of loop 
Angle between magnetic field and area vector 
21°
Form the formula of torque in case of magnetic field,
г 
Where 
magnetic moment
г 
г 
г 
Nm
Therefore, the torque on the loop is Nm
