Answer:
Induced emf in the coil, E = 0.157 volts
Explanation:
It is given that,
Number of turns, N = 100
Diameter of the coil, d = 3 cm = 0.03 m
Radius of the coil, r = 0.015 m
A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.
Due to this change in magnetic field, an emf is induced in the coil which is given by :


E = -0.157 volts
Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.
Answer:
40 N
Explanation:
F=ma where F is the applied force, m is the mass of object and a is the acceleration.
Since there is no friction, substituting 20 Kg for m and 2 m/s squared for a then we obtain
F=20*2=40 N
Answer:
E = 0.18 J
Explanation:
given,
Potential of the battery,V = 9 V
Charge on the circuit, Q = 20 m C
= 20 x 10⁻³ C
energy delivered in the circuit
E = Q V
E = 20 x 10⁻³ x 9
E = 180 x 10⁻³
E = 0.18 J
Energy delivered in the circuit is equal to E = 0.18 J
An electric power measure the rate of electrical energy transfer by an electric circuit per unit of time.