A copper ring has a radius of 4.0 cm and a resistance of 1 mΩ. A magnetic field is applied over the ring, perpendicular to its plane. If the magnetic field increases from 0.2 T to 0.4 T in a time interval of 5 x10⁻³ s. What is the current in the ring? (Ans: 201 A).
Data Given:
Number of turns of the ring = N = 1
Radius of copper ring = r = 4.0 cm = 0.04 m
Resistance = R = 1 mΩ = 1x10⁻³ Ω
1st Magnetic Field = B₁ = 0.2 T
1st Magnetic Field = B₂ = 0.4 T
ΔB = B₂ - B₁ = 0.4 T -0.2 T = 0.2 T
Time interval = △t = 5x10⁻³ s
To Find:
Current in the ring = `\I` = ?
Solution:
Using the following relation for `\I`
`\I` = `\frac {Ɛ}{R}` ---------------(1)
Ɛ = -N `\frac {Δф}{Δt}`
where Δф = ΔB . A = ΔB A cos θ , A = ㄫ r²
taking absolute value the negative sign will become positive
|Ɛ| = |-N `\frac {ΔB ㄫ r²}{Δt}`|
Ɛ = N `\frac {ΔB ㄫ r²}{Δt}`
by putting the corresponding values
Ɛ = 1 x `\frac {0.2 T x 3.1416 x (0.04 m)² }{5x10⁻³ s}`
Ɛ = `\frac {0.62832 T x 0.0016 m²}{5x10⁻³ s}`
Ɛ = `\frac {0.001005 T m²}{5x10⁻³ s}`
Ɛ = 0.0002010624 x10³ V
or
Ɛ = 0.201 V
Now putting the corresponding values in equation (1)
`\I` = `\frac {0.20106 V}{ 1x10⁻³ Ω}`
`\I` = 0.20106 x10³ A
or
`\I` = 201 A ------------Ans.
************************************
************************************
Shortcut Links For
1. Website for School and College Level Physics 2. Website for School and College Level Mathematics 3. Website for Single National Curriculum Pakistan - All Subjects Notes
© 2022-Onwards by Academic Skills and Knowledge (ASK)
Note: Write me in the comment box below for any query and also Share this information with your class-fellows and friends.
1. Website for School and College Level Physics
2. Website for School and College Level Mathematics
3. Website for Single National Curriculum Pakistan - All Subjects Notes
© 2022-Onwards by Academic Skills and Knowledge (ASK)
Note: Write me in the comment box below for any query and also Share this information with your class-fellows and friends.
0 Comments
If you have any QUESTIONs or DOUBTS, Please! let me know in the comments box or by WhatsApp 03339719149