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)

but the value emf "Ɛ" is unknown, so let it be find first by using the following relation

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.


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