First slide

Magnetic flux and induced emf

Question

A conductor of length I and mass m can slide without any friction along the two vertical conductors connected at the top through a capacitor (figure). A uniform magnetic field B is setup $⊥$ to the plane of paper. The acceleration of the conductor

Difficult

A

is constant
B

increases
C

decreases
D

cannot say

Solution

Let v is the velocity of conductor at any time,
then induced emf : $e = B l v$ ……..(i)
Charge on capacitor : $q = C e = C B l v$
Current in circuit : $i = \frac{d q}{d t} = C B l \frac{d v}{d t}$
for conductor: $m g - i B l = \frac{m d v}{d t}$
$\Rightarrow m g - C B^{2} l^{2} \frac{d v}{d t} = \frac{m d v}{d t} \Rightarrow \frac{d v}{d t} = \frac{m g}{m + C B^{2} l^{2}}$
This is the acceleration of conductor which is constant.

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