The coil of an a.c. generator has 100 turns, each of cross-sectional area 2 ~m^2 . It is rotating at constant angular speed 30 rad / s , in a uniform magnetic field of 2 × 10^ -2 ~T . If the total resistance of the

Correct answer: C. 12\ W aligned & N=100, ~A=2 ~m^2, ω=30 rad / s & B=2 × 10^ -2 ~T, R=600 aligned Maximum power dissipated in the circuit aligned P_ =E_ r m s × I_ r m s & = E_0 2 × I_0 2 & =(E_0 I_0)/(2)...(i) aligned But I_0= E_0 R ...(ii) Putting (2) into (1) we get, I_0= E_0 ^2 2 R aligned But E_0 & =NAB ω E_0 & =100 × 2 × 2 × 10^ -2 × 30 & =120 ~V P_ = & (120 × 120)/(2 × 600)=12 ~W aligned

JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Magnetic Properties of Matter

JEE Main · Physics · Magnetic Properties of Matter

The coil of an a.c. generator has 100 turns, each of...

The coil of an a.c. generator has 100 turns, each of cross-sectional area

2 \mathrm{~m}^2

. It is rotating at constant angular speed

30 \mathrm{rad} / \mathrm{s}

, in a uniform magnetic field of

2 \times 10^{-2} \mathrm{~T}

. If the total resistance of the circuit is

600 \Omega

then maximum power dissipated in the circuit is

Options

A $6\ W$
B $9\ W$
C $12\ W$
D $24\ W$

Correct answer

C. $12\ W$

Step-by-step solution

\begin{aligned} & \mathrm{N}=100, \mathrm{~A}=2 \mathrm{~m}^2, \omega=30 \mathrm{rad} / \mathrm{s} \\ & \mathrm{B}=2 \times 10^{-2} \mathrm{~T}, \mathrm{R}=600 \Omega \end{aligned}

Maximum power dissipated in the circuit

\begin{aligned} P_{\max }=E_{r m s} \times I_{r m s} & =\frac{E_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \\ & =\frac{E_0 I_0}{2}...(i) \end{aligned}

But

\mathrm{I}_0=\frac{\mathrm{E}_0}{\mathrm{R}}...(ii)

Putting (2) into (1) we get,

\mathrm{I}_0=\frac{\mathrm{E}_0{ }^2}{2 \mathrm{R}}

\begin{aligned} \text { But } \mathrm{E}_0 & =\mathrm{NAB} \omega \\ \mathrm{E}_0 & =100 \times 2 \times 2 \times 10^{-2} \times 30 \\ & =120 \mathrm{~V} \\ \therefore \quad \mathrm{P}_{\max }= & \frac{120 \times 120}{2 \times 600}=12 \mathrm{~W} \end{aligned}

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