Solved Numerical Problem No. 15.12 ( Unit-15: Electromagnetic Induction); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

A generator has a rectangular coil consisting of 360 turns. The coil rotates at 420 revs per min in a 0.14 T magnetic field. The peak value of emf produced by the generator is 50 V. If the coil is 5 cm wide, find the length of the side of the coil. (Ans: 45 cm). 

Data Given:

Number of turns = N = 360 

Angular speed = ധ = 420 revs per min = `\frac {420 ｘ2π rad}{60s}` =  `\frac {420 ｘ2 ｘ3.1416 rad}{60s}` = 43.98 rad s⁻¹

Magnetic field strength = B = 0.14 T

Peak value emf = `\Ɛ_0` = 50 V

Width of the side of Rectangular coil = w = 5 cm = 0.05 m

To Find:

Length of the rectangular coil = `\l` = ?

Solution:

The relation for the Peak value emf  `\Ɛ_0` is given 

`\Ɛ_0`  = B ധ N A

where Area = A = length ｘ Width = `\l` ｘ w

`\Ɛ_0`  = B ധ N `\l` w

or

`\l` = `\frac {Ɛ_0}{BNധw}`

Putting the corresponding values 

`\l` = `\frac {50 V}{0.14 T ｘ 360 ｘ 43.98 rad s⁻¹ ｘ 0.05 m}`

`\l` = `\frac {50 V}{110.8296 T rad s⁻¹ m}`

`\l`  = 0.451 m

or

`\l`  = 45.1 cm

or

`\l` = 45 cm  -------------Ans.

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