Question

A magnet is pushed into a solenoid of 20 coils with a cross-sectional area of 4.0x10-4 m2. As a result, the magnetic field inside the coils changes at a uniform rate from 0.0500 T to 0.180 T in a time interval of 1.90 s.1. What is the magnitude of the emf in [mV] generated in the solenoid?2. If a 5 ohm resistor is connected to the solenoid, what is the magnitude of the current flowing through it in [mA]?

Answer 1

Given:

The number of the coils, N=20

The cross-sectional area of the solenoid, A=4.0×10⁻⁴ m²

The magnetic fields, B₁=0.0500 T

and B₂=0.180 T

The time interval, t=1.90 s

The resistance of the resistor, R=1.90 s

To find:

1. The magnitude of the induced emf.

2. The current flowing through the resistor.

Explanation:

1.

The magnitude of the induced emf in the solenoid is given by,

$E=NA\frac{(B_2-B_1)}{t}$

On substituting the known values,

$\begin{gathered} E=20\times4.0\times10^{-4}\times\frac{(0.180-0.0500)}{1.90} \\ =0.55\times10^{-3}\text{ V} \\ =0.55\text{ mV} \end{gathered}$

2.

From Ohm's law, The induced emf is given by,

$E=IR$

On substituting the known values,

$\begin{gathered} 0.55\times10^{-3}=I\times5 \\ \Rightarrow I=\frac{0.55\times10^{-3}}{5} \\ =0.11\times10^{-3}\text{ A} \\ =0.11\text{ mA} \end{gathered}$

Final answer:

1. The induced emf is 0.55 mV

2. The current through the given resistor is 0.11 mA