Answer:
9.44 eV
Explanation:
q = Amount of charge = 86.3 μC = 86.3 x 10⁻⁶ C (Since 1 μC = 10⁻⁶ C)
V = Electric potential of the charge = 17.5 Volts
U = Amount of electric potential energy carried by charge
Amount of electric potential energy carried by charge is given as
U = q V
Inserting the values
U = (86.3 x 10⁻⁶) (17.5)
U = 0.00151 J
U = 
U = 9.44 eV
KE = Kinetic energy the charge could have
Using conservation of energy
KE = U
KE = 9.44 eV
Answer:
e = 10 V
Explanation:
given,
number of the coaxial loops = 10
Cross sectional area = 0.5 m²
magnitude of magnetic field =
B = 3 T + (2 T/s)*t.
B = ( 3+ 2 t ) T
induced potential difference = ?
At time = 2 s
we know,
induced emf

∅ = B . A




e = -10 V
magnitude of induced emf
|e| = |-10 V|
e = 10 V
the induced potential difference in the loop = e = 10 V