In the given circuit, inductor of $L = 1 m H$ and resistance $R = 1 Ω$ are connected in series to the ends of two parallel conducting rods as shown. Now a rod of length $10 c m$ is moved with constant velocity of $1 c m / s$ in magnetic field $B = 1 T$ . If rod starts moving at $t = 0$ , then the current in circuit after 1 millisecond is $x \times 10^{3} A$ . Then the value of $x$ is (given $e^{- 1} = 0.37$ )

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Solution

If this rod will move then there will be a change in magnetic flux and if magnetic flux will change then emf will develop.
So,
$e m f = B v l$ , where $v$ is the velocity of rod
$E = 1 \times 10^{- 2} \times 10^{- 1} = 10^{- 3} V$
This will behave as LR circuit as shown in figure
As we switch on the circuit there will be growth of current in it.
$i = i_{m a x} (1 - e^{\frac{- t R}{L}})$
$i_{m a x} = \frac{E}{R} = \frac{10^{- 3}}{1} = 10^{- 3} A$
$i_{m a x} = 10^{- 3} (1 - e^{\frac{- 10^{- 3} \times 1}{10^{- 3}}})$
$i_{m a x} = 10^{- 3} (1 - e^{- 1}) = 10^{- 3} (1 - 0.37)$
$i_{m a x} = 0.63 \times 10^{- 3} A$
so, $x = 0.63$

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A $10 c m$ long perfectly conducting wire PQ is moving with a velocity $1 c m / s$ on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor $L = 1 m H$ and a resistance $R = 1 Ω$ as shown in figure. The horizontal rail, L and R lie in the same plane with a uniform magnetic field $B = 1 T$ perpendicular to the plane. If the key $S$ is closed at certain instant , the current in the circuit after 1 millisecond is $x \times 10^{- 3} A$ , where the value of $x$ is .
[Assume the velocity of wire PQ remains constant $(1 c m / s)$ after key S is closed. Given : $e^{- 1} = 0.37$ , where e is base of the natural logarithm]