SATHEE: electromagnetic-induction Question 15

इलेक्ट्रोमैग्नेटिक इंडक्शन

Question: Q. 4. The magnetic field through a single loop of wire, $12 cm$ in radius and $8.5 Ω$ resistance, changes with time as shown in the figure. The magnetic field is perpendicular to the plane of the loop. Plot induced current as a function of time.

A [CBSE SQP 2015]

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Solution:

Ans.

$\begin{aligned} ε & = - \frac{d ϕ}{d t} & = - π R^{2} \times \frac{d B}{d t} & = - \frac{22}{7} \times (0.12)^{2} \times \frac{1}{2} ε & = - 0.023 V I & = \frac{ε}{R} & = - 2.7 mA for 0 < t < 2 s . \end{aligned}$

Similarly :

	$0 < t < 2 s$	$2 < t < 4 s$	$4 < t < 6 s$
$ε (V)$	-0.023	0	+0.023
$I (mA)$	-2.7	0	+2.7

[AI Q. 5. A rectangular conductor $L M N O$ is placed in a uniform magnetic field of $0.5 T$ . The field is directed perpendicular to the plane of the conductor.

When the arm $M N$ of length of $20 cm$ is moved towards left with a velocity of $10 {ms}^{- 1}$ , calculate the emf induced in the arm. Given the resistance of the arm to be $5 Ω$ (assuming that other arms are of negligible resistance) find the value of the current in the arm.

A [O.D. I, II, III 2013]

Ans. Let $O N$ be at some point $x$ .

The emf induced in the loop $= ε$

$\begin{aligned} ε & = \frac{- d ϕ}{d t} = \frac{- d (B l x)}{d t} = B l v & = 0.5 \times 0.2 \times 10 = 1 V \end{aligned}$

$∴$ Current in the arm,

$\begin{matrix} (1) & I = \frac{ε}{R} = \frac{1}{5} = 0.2 A \end{matrix}$

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इलेक्ट्रोमैग्नेटिक इंडक्शन

विषयसूची

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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