A large long solenoid is made with 60 turns/cm. A smaller solenoid has 30 turns,

Question

A large long solenoid is made with 60 turns/cm. A smaller solenoid has 30 turns, is 2.0 cm long and has a diameter of 0.80 cm. The ends of the small solenoid are connected to a 10 ohm resistor. The small solenoid is placed right in the centre of the large solenoid, Fig. 2. In the drawing the small solenoid is shown outside the large solenoid for clarity. The large solenoid is then connected to a DC power supply which changes the current in the large solenoid from zero to 10 A in a time of 2.0 ms. Calculate the magnitude of the magnetic field in the large solenoid when the current is at its maximum value of 10 A.

Explanation / Answer

I guess you area asking only question (2). So I am solving it:

2)

Magnetic field inside the long solenoid, B = u*n*I

where u = permeability of air = 1.26*10-6

n = turns/meter = 60/cm = 6000/m

I = current

So, the flux through the smaller solenoid,

Q = B*A

where A = area of the smaller solenoid = pi*r^2

r = 0.8/2 cm = 0.004 m

Induced emf when the magnetic field is change is given by :

E = dQ/dt = A*dB/dt = A*(u*n*dI/dt)

where dI = change in current = 10 -0 = 10 A

dt = time duration = 2 ms = 0.002 s

Magnitude of magnetic field in the large solenoid when current = I = 10 A,

B = unI = 1.26*10^-6*6000*10 = 0.076 T <-----------answer

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