A pair of parallel conducting rails lie at right angle to a uniform magnetic field of 2.0 T as shown in the fig. Two resistors 10 Ω and 5 Ω are to slide without friction along the rail. The distance between the conducting rails is 0.1 m. Then Options: A . &nbspInduced current = 1150A directed clockwise if 10Ω resistor is pulled to the right with speed 0.5ms−1 and 5Ω resistor is held fixed B . &nbspInduced current = 1300A directed anti-clockwise if 10Ω resistor is pulled to the right with speed 0.5ms−1 and 5Ω resistor is held fixed C . &nbspInduced current = 1300A directed clockwise if 5Ω resistor is pulled to the left at speed 0.5ms−1 and 10Ω resistor is held at rest D . &nbspInduced current =1150A directed anti-clockwise if 5Ω resistor is pulled to the left with speed 0.5ms−1 and 10Ω resistor is held at rest

A pair of parallel conducting rails lie at right angle to a uniform magnetic fie

Question

A pair of parallel conducting rails lie at right angle to a uniform magnetic field of 2.0 T as shown in the fig. Two resistors 10

Ω

and 5

Ω

are to slide without friction along the rail. The distance between the conducting rails is 0.1 m. Then

A Pair Of Parallel Conducting Rails Lie At Right Angle To A ...

Options:

A . Induced current = 1150A directed clockwise if 10Ω resistor is pulled to the right with speed 0.5ms−1 and 5Ω resistor is held fixed

B . Induced current = 1300A directed anti-clockwise if 10Ω resistor is pulled to the right with speed 0.5ms−1 and 5Ω resistor is held fixed

C . Induced current = 1300A directed clockwise if 5Ω resistor is pulled to the left at speed 0.5ms−1 and 10Ω resistor is held at rest

D . Induced current =1150A directed anti-clockwise if 5Ω resistor is pulled to the left with speed 0.5ms−1 and 10Ω resistor is held at rest

Answer: Option D
:
D
When 5

Ω

resistor is pulled left at 0.5 m/sec induced emf., in the said resistor

= e = v B l = 0.5 \times 2 \times 0.1 = 0.1

V
Resistor

10 Ω

is at rest so induced emf in it (e=vBl) be zero.
Now net emf., in the circuit=0.1V and Equivalent resistance of the circuit
R=

15 Ω

Hence current

i = \frac{0.1}{15}

amp=

\frac{1}{150}

amp
And its direction will be anti-clockwise (according to Lenz's law)

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