Electromagnetic Induction And Alternating Current Ques 1
1 The figure shows a square loop $L$ of side $5 \mathrm{~cm}$ which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of $1 \mathrm{~cm} \mathrm{~s}^{-1}$. At some instant, a part of $L$ is in a uniform magnetic field of $1 \mathrm{~T}$, perpendicular to the plane of the loop. If the resistance of $L$ is $1.7 \Omega$, the current in the loop at that instant will be close to
(Main 2019, 12 April I)
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Answer:
Correct Answer: 1.( b )
Solution:
- Induced emf in the conductor of length $L$ moving with velocity of $1 \mathrm{~cm} / \mathrm{s}$ in the magnetic field of $1 \mathrm{~T}$ is given by
$ V=B L v $
If equivalent resistance of the circuit is $R_{\mathrm{eq}}$, then current in the loop will be
$ i=\frac{V}{R_{\mathrm{eq}}}=\frac{B L v}{R_{\mathrm{eq}}} $
Now, given network is a balanced Wheatstone bridge $\left(\frac{P}{Q}=\frac{R}{S}\right)$.
$ R_W=\frac{2 \times 4}{2+4}=\frac{8}{6}=\frac{4}{3} \Omega $
Again, resistance of conductor is $1.7 \Omega$.
So, effective resistance will be
$ \begin{aligned} & R_{\text {eq }}=\frac{4}{3}+1.7=\frac{4}{3}+\frac{17}{10} \\ & R_{\text {eq }}=\frac{40+51}{30}=\frac{91}{30} \simeq 3 \Omega \end{aligned} $
By putting given values of $R_{\text {eq }}, B$ and $v$ in Eq. (ii), we have
$ i=\frac{(1)\left(5 \times 10^{-2}\right) \times 10^{-2}}{3} $
here, $L$
$ \begin{aligned} & i=\frac{5 \times 10^{-4}}{3}=1.67 \times 10^{-4} \mathrm{~A} \\ & i=167 \mu \mathrm{A}=170 \mu \mathrm{A} \end{aligned} $
BETA
