Question

A solenoid runs though the x-y plane with 3/m x 10⁴ (turns per meter) with current that is always counter-clockwise. Initially, the current is 5 amps for the solenoid. A square wire with sides of 0.49 meters is wrapped 21 times and is placed in the x-y plane within the solenoid. Due to the current changing in the solenoid over a time of 0.30 seconds, the average induced current creates a magnetic moment of 73.7 x 10-3 A m2 in the -z direction. If the resistance of the square wire is 39 Ohms, what is the final current of the solenoid in amps?

Answer 1

n = number of turns per unit length in solenoid = 3*10⁴ turns per meter

i_o = initial current = 5 A

i_f = final current = ?

t = time = 0.30 sec

Change in magnetic field is given as

Delta B = u_o*n*(i_f - i_o) --------------------------------------> [eqn -1]

a = length of side of square = 0.49 m

A = area of the square loop = a² = (0.49)² = 0.24 m²

N = number of turns in square loop = 21

t = time = 0.30 sec

induced emf is given as

E = N*Delta B*A/t

R = resistance of square wire = 39 ohms

induced current is given as

i = E/R = N*Delta B*A/(Rt)--------------------------------> [eqn - 2]

magnetic moment, 'T' is given as

T = N*i*A

0.0737 = 21 (0.24) i

i = 0.015 A

using [eqn - 2]

i = N*Delta B*A/(Rt)

0.015 = (21)*Delta B *(0.24)/(39*0.30)

Delta B = 0.035 T

using [eqn - 1]

Delta B = u_o*n*(i_f - i_o)  

0.035 = (12.56*10^-7)*(3*10⁴) (i_f - 5)

Final Current of the Solenoid, i_f = 5.93 A