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 = ƐR ---------------(1)
Ɛ = -N ΔфΔt
where Δф = ΔB . A = ΔB A cos θ , A = ㄫ r²
taking absolute value the negative sign will become positive
|Ɛ| = |-N ΔBㄫr²Δt|
Ɛ = N ΔBㄫr²Δt
by putting the corresponding values
Ɛ = 1 x 0.2Tx3.1416x(0.04m)²5x10⁻³s
Ɛ = 0.62832Tx0.0016m²5x10⁻³s
Ɛ = 0.001005Tm²5x10⁻³s
Ɛ = 0.0002010624 x10³ V
or
Ɛ = 0.201 V
Now putting the corresponding values in equation (1)
I = 0.20106V1x10⁻³Ω
I = 0.20106 x10³ A
or
I = 201 A ------------Ans.
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