Question

A light bulb and a solenoid are connected in series to a battery. An irod rod is thrust rapidly into the solenoid and later rapidly removed. The bulb 1. No change in, Dims out 2. Brightens in, No change out 3. Dims in, Brightens out 4. No change in, Brightens out 5. Dims in, Dims out 6. Brightens in, Brightens out 7. Brightens in, Dims out 8. Remains brighter the entire time 9. No change in or out 10. Dims in, No change out 006 10.0 poi

Answer 1

Answer:

3) Dims in, Brightens out

Explanation:

The relationship between the inductance of a coil and the current is inverse.

As the rod is thrusted into the solenoid, the inductance increases, which in turn decreases the current and the bulb gets dimmer. This is also because some magnetic force is drawn away from the solenoid and the battery.

When the rod is removed, the inductance of the solenoid decreases, the current increases, and the bulb becomes brighter.

Answer 2

Answer:

The correct option is;

3. Dims in, Brightens out

Explanation:

Inserting the iron rod into the current carrying solenoid causes the inductance and hence impedance to increase. Increasing impedance increases the circuit total resistance to current flow, thereby reducing the amount of  electric current, which causes the light bulb to dim as the iron rod enters the solenoid.

The reverse happens as the iron rod is rapidly removed.

The inductive reactance is given by;

Inductive reactance, $X_L = \omega L$ = 2·π·f·L

Where;

L = Inductance (Henries)

ω = Angular velocity in radians

f = Frequency

current, I is given by

$I = \frac{Voltage }{Resistance} = \frac{V}{X_L}$

Therefore

$I\propto\frac{1}{X_L}$$I \propto\frac{1}{X_L}$

Iron rod entering the solenoid increases inductance, decreases the current and the bulb is dimmer.