A 10 cm long perfectly conducting wire PQ is moving with a velocity 1 cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L=1 mH and a resistance R=1Ω 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×10−3 A, where the value of x is
[Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given : e−1=0.37, where e is base of the natural logarithm]
A 10 cm long perfectly conducting wire PQ is moving with a velocity 1 cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L=1 mH and a resistance R=1Ω 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×10−3 A, where the value of x is
[Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given : e−1=0.37, where e is base of the natural logarithm]
Here when the rod starts moving, its going to change the magnetic flux and hence emf will be devoloped.
The equivalent circiut in a simplified form would look like this,
ε=(→v×→B)→ℓ=Bvℓ=10−2×1×10−1
ε=10−3 volt
i=εR(1−e−RtL)=10−31(1−e−1)
i=10−3(1−0.37)
I=0.63mA
Here when the rod starts moving, its going to change the magnetic flux and hence emf will be devoloped.
The equivalent circiut in a simplified form would look like this,
ε=(→v×→B)→ℓ=Bvℓ=10−2×1×10−1
ε=10−3 volt
i=εR(1−e−RtL)=10−31(1−e−1)
i=10−3(1−0.37)
I=0.63mA
