A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed normally to the loop. What is the e.m.f. developed across the cut if the velocity of the loop is 1 cm s⁻¹ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? (Answer: 2.4 ｘ10⁻⁴ V, 2 s, & 0.6 ｘ 10⁻⁴ V, 8 s)  



Data Given:

Length of the wire loop = L = 8 cm = 0.08 m

Width of the wire loop = W = 2 cm = 0.02 m

Magnetic field = B = 0.3 T

Velocity of the loop = v = 1 cm s⁻¹  = 0.01 m s⁻¹



To Find:

(a) Induced e.m.f. in the longer side = `\Ɛ_L` = ?

(b) Induced e.m.f. in the shorter side = `\Ɛ_S` = ?

(c) Time taken by the induced voltage in each case i.e
     `\t_L` = ? and  `\t_S` = ?


Solution:

(a) The equation for the motional e.m.f. for longer side is given by:

`\Ɛ_L`  = B l v

by putting values

`\Ɛ_L`  = 0.3 T ｘ 0.08 m ｘ 0.01 m s⁻¹

`\Ɛ_L` = 0.00024 V

`\Ɛ_L`  = 2.4 ｘ 10⁻⁴ V ------------Ans. (1)


(b) The equation for the motional e.m.f. for the longer side is given by:

`\Ɛ_S` = B W v

by putting values

`\Ɛ_S`   = 0.3 T ｘ 0.02 m ｘ 0.01 m s⁻¹

`\Ɛ_S`   = 0.00006 V

`\Ɛ_S`   = 0.6 ｘ 10⁻⁴ V ------------Ans. (2)


(c) Time taken by the induced voltage in each case i.e. `\t_L` = ? and  `\t_S` = ?

The general Formula for finding time is 

t = `\frac {S}{v}`               ∴[ S = v t}

Since the longer side moves along the width so the time taken by induced voltage for the longer side is 

`\t_L` = `\frac {W}{v}`

putting values

`\t_L` = `\frac {0.02 m}{0.01 m s⁻¹}`

`\t_L` = 2 s ----------------Ans. 

And the shorter side moves along the length so the time taken by induced voltage for the shorter side is 

`\t_S` = `\frac {L}{v}`

putting values

`\t_S` = `\frac {0.08 m}{0.01 m s⁻¹}`

`\t_S` = 8 s ----------------Ans. 


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