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Question
Consider the situation shown in figure. The wire PQ has a negligible resistance and is made to slide on the three rails with a constant speed of 5 cm s−1. Find the current in the 10 Ω resistor when the switch S is thrown to (a) the middle rail (b) the bottom rail.
Solution
Given:-
Magnetic field, B = 1 T
Velocity of the sliding wire, v = 5 × 10−2 m/s
Resistance of the connected resistor, R = 10 Ω
(a) When the switch is thrown to the middle rail:-
Length of the sliding wire = 2 × 10−2 m
Induced emf, E = Bvl
= 1 × (5 × 10−2) × (2 × 10−2) V
= 10 × 10−4 = 10−3 V
Current in the 10 Ω resistor is given by
\[i = \frac{E}{R}\]
\[ = \frac{{10}^{- 3}}{10} = {10}^{- 4} = 0 . 1 mA\]
(b) When the switch is thrown to the bottom rail:-
The length of the sliding wire becomes 4 × 10−2 m.
The induced emf is given by
E = Bvl'
= 1 × (5 × 10−2) × (4 × 10−2)
= 20 × 10−4 V
Now,
Current, \[i = \frac{20 \times {10}^{- 4}}{10}A\]
= 2 × 10−4 A = 0.2 mA
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