3 InductionCoilSolenoidA closed copper coil with 100 turns of wire and a total resistance of R = 5.00 is placed outside of a solenoid.• The radius of the coil is a = 2.0 cm.• The solenoid is 10 cm long and has 2,000 turns of wire. Its radius is a = 1.5 cm.• The solenoid is connected to a power supply and the current increases steadily from 0 to 2.0 A in 50 ms.What current is induced in the copper coil?

Given Data The number of turns of copper coil is:N=100 turns The total resistance is:R=5Ω The…

Answered: 3 Induction Coil Solenoid A closed…

3 Induction Coil Solenoid A closed copper coil with 100 turns of wire and a total resistance of R = 5.00 is placed outside of a solenc • The radius of the coil is a = 2.0 cm. 0.000

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Question

3 Induction
Coil
Solenoid
A closed copper coil with 100 turns of wire and a total resistance of R = 5.00 is placed outside of a solenoid.
• The radius of the coil is a = 2.0 cm.
• The solenoid is 10 cm long and has 2,000 turns of wire. Its radius is a = 1.5 cm.
• The solenoid is connected to a power supply and the current increases steadily from 0 to 2.0 A in 50 ms.
What current is induced in the copper coil?

Expert Solution

Step 1

Given Data

The number of turns of copper coil is: $N = 100 turns$
The total resistance is: $R = 5 Ω$
The radius of the coil is: $a = 2 cm = 2 \times 10^{- 2} m$
The length of solenoid is: $L = 10 cm = 0.1 m$
The number of turns of solenoid is: $n = 2000 turns$
The radius of solenoid is: $a = 1.5 cm = 0.015 m$
The change in current is: $∆ I = 2 A$
The change in time is $∆ t = 50 ms = 50 \times 10^{- 3} s$

To find:

The current is induced in the copper coil,

Formula used:

The magnetic field due to solenoid is given as,

$B = μ_{o} n I$

Here,

B is the magnetic field

$μ_{o}$ is the magnetic permiability

$n$ is the number of turns of solenoid.

I is the current

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