A Ruhmkorff coil is made by wrapping a solenoid around a second solenoid as show

Question

A Ruhmkorff coil is made by wrapping a solenoid around a second solenoid as shown in the figure. Assume that the long solenoid has a length of L1 0.135 m, radius of R1 0.037 m and is made up of 2000 turns the short solenoid has a length of L2 0.090 m, radius of R2 0.010 m and is made up of 200 turns. A current I 0.3 A flows in the long solenoid in the direction shown by the arrow a) What is in Teslas the magnitude of the magnetic field produced inside the long solenoid? SUBMIT b) What would be in Webers the magnetic flux in one loop of the short solenoid due to the magnetic field generated by the long solenoid? SUBMIT C) What would be in volts the magnitude of the induced voltage in the short solenoid if the direction of the current in the long solenoid is flipped in 0.03 seconds? SUBMIT d) hat is the direction of flow of the induced current (think of the direction to the wire to right of the solenoid, the wire that is connecting the voltmeter note for example that l is pointing up)? down up SUBMIT e) What would be in volts the magnitude of the induced voltage in the short solenoid if the current l in the long solenoid is instead increased to 0.5 A in 0.03 seconds? SUBMIT

Explanation / Answer

here,

l1 = 0.135 m

R1 = 0.037 m

no.of turns on solenoid long , n1 = 2000

l2 = 0.090 m

R2 = 0.010 m

no.of turns on short solenoid , n2 = 200

current , I = 0.3 A

a)

the magnitude of magnetic feild , B = u0 * n1 * I / (l1)

B = u0 * 0.3 * 2000 / (0.135)

B = 5.59 * 10^-3 T

the magnitude of magnetic feild is 5.59 * 10^-3 T

b)

the magnitude of magnetic flux in one loop of short solenoid due to magnetic feild generated in the long solenoid , E = B * Area of short solenoid

E = 5.59 * 10^-3 * pi * (0.010)^2

E = 1.75*10^-6 weber

the magnitude of magnetic flux in one loop of short solenoid due to magnetic feild generated in the long solenoid is 1.75*10^-6 weber

c)

the magnitude of the induced voltage in the short solenoid be V

V = 2 * n2 * (magnetic flux) / time

V = 2 * 200 * 1.75*10^-6 / 0.03

V = 0.0233 V

the magnitude of the induced voltage in the short solenoid is 0.0233 V

d)

using lenz's law,

the direction of flow of induced current is up

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