A circular coil has 15 turns of radius 2cm each. The plane of the coil is inclined at 40° to a uniform magnetic field of 0.2 T. If the field is increased to 0.5 T in 0.2 sec, find the magnitude of the induced emf. (Answer: 1.8 ｘ10⁻² V). 



Data Given:

Number of turns of the coil = N = 15 turn 

Radius of each coil = r = 2 cm = 0.02 m

Angle between the coil and B = θ = 40° 

1st Magnetic Field = B₁ = 0.2 T

1st Magnetic Field = B₂ = 0.5 T

ΔB = B₂ - B₁ = 0.5 T -0.2 T = 0.3 T

Time interval = △t = 0.2 s


To Find:


Induced e.m.f. in the coil = Ɛ = ?


Solution:

Using the following relation

Ɛ = -N `\frac {Δф}{Δt}`

where Δф = ΔB . A = ΔB A cos θ  , A = ㄫ r² and angle θ between the B and A will be θ = 90° - 40° = 50°

taking absolute value the negative sign will become positive

|Ɛ| = |-N  `\frac {ΔB ㄫ r² cos θ}{Δt}`|

Ɛ = N  `\frac {ΔB ㄫ r² cos θ}{Δt}`

by putting the corresponding values

Ɛ = 15 ｘ `\frac {0.3 T ｘ 3.1416 ｘ (0.02 m)² ｘ cos 50° }{0.2 s}`

Ɛ = `\frac {14.137 T ｘ 0.0004 m² ｘ 0.643}{0.2 s}`

Ɛ = `\frac {0.00364 T m²}{0.2 s}`

Ɛ = 0.01818

or


Ɛ =  1.818  ｘ 10⁻² V ---------------Ans.





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