A solenoid 2.0 m long and 30 cm in diameter consists of 5000 turns of wire (prim

Question

A solenoid 2.0 m long and 30 cm in diameter consists of 5000 turns of wire (primary). A 5 m coil (secondary) with negligible resistance is wrapped around the solenoid and connected to a 180 ohm resistor. The solenoid current is given by I = I0 sin wt where I0= 85 A, w=210 rad/s, and t is in seconds.

(1) Determine the magnetic field at the center of the solenoid (primary) at t=1 ms.

(2) Determine the peak emf induced in the secondary loop.

(3) What is the current in the secondary coil when the primary coil current is a maximum?

Explanation / Answer

a) The magnetic field of the solenoid is

B = ( Ns 0 i ) / L

B = ( Ns 0 I0 sin(t) ) / L

B = ( 5000 * 4x10-7 T.m/A * 85 A * sin(t) ) / ( 2.0 m )

B = ( 0.267 T ) * sin(t)

hence, at t = 1ms

B = ( 0.267 T ) * sin(210 rad/s * 1x10-3 s)

B = 0.056 T

b) The induced emf on the coil is

= -NcdB / dt

where

B = A B

B = r2 B

B = * ( 0.15 m )2 * ( 0.267 T.m ) * sin(t)

B = ( 0.019 T.m2 ) sin(t)

hence,

= -Ncd[ ( 0.019 T.m2 ) sin(t) ] / dt

= -Nc * ( 0.019 T.m2 ) * cos(t)

the peak of efm is obtained when cos(t)=1

peak = - 5 * ( 0.019 T.m2 ) * 210 rad/s

peak = - 19.9 v

c) when current in solenoid is maximum the emf is maximum too,, hence

the current in the coil is

i =  peak / R

i = ( -19.9 ) / ( 180 )

i = -0.11 A

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