SATHEE: Electro Magnetic Induction And Alternating Currents Question 475

Electro Magnetic Induction And Alternating Currents Question 475

Question: A plane loop, shaped as two squares of sides a =1 m and b=0.4 m is introduced into a uniform magnetic field $⊥$ to the plane of loop. The magnetic field varies as $B = 10^{- 3} \sin$ (100t) T. The amplitude of the current induced in the loop if its resistance per unit length is $r = 5, m Ω^{- 1}$ is

Options:

A) 2 A

B) 3 A

C) 4 A

D) 5 A

Show Answer

Answer:

Correct Answer: B

Solution:

$ϕ$ (flux linked) $= a^{2} B, \cos, 0^{o} - b^{2} B, \cos, 180^{o}$

$E = - \frac{d ϕ}{d t} = - (a^{2} - b^{2}) \frac{d B}{d t}$

$= (a^{2} - b^{2}) B_{0}, ω \cos, ω t$ where $B = B_{0}$ , $\sin, ω t$ ,

$B_{0} = 10^{- 3} T$ ,

$ω = 100$
$∴ I_{max} = (a^{2} - b^{2}) \frac{B_{0} ω}{R}$

and $R = (4 a + 4 b) r = 4 (a + b) r$
$∴ I_{max} = \frac{(a - b) B_{0} ω}{4 r} = \frac{(1 - 0.4) \times 10^{- 3} \times 100}{4 \times 5 \times 10^{- 3}} = 3 A$

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Electro Magnetic Induction And Alternating Currents Question 475

Question: A plane loop, shaped as two squares of sides a =1 m and b=0.4 m is introduced into a uniform magnetic field ⊥ to the plane of loop. The magnetic field varies as B=10−3sin (100t) T. The amplitude of the current induced in the loop if its resistance per unit length is r=5,mΩ−1 is

Options:

Answer:

Solution:

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