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A conducting bar mass m and length l mov...

A conducting bar mass `m` and length `l` moves on two frictionless parallel conducting rails in the presence of a uniform magnetic field `B_(0)` directed into the paper. The bar is given an initial velocity `v_(0)` to the right and is released at `t = 0`. The velocity of the bar as a function of time is given by

A

`v_(0) xx (mRt)/(B_(0)^(2)l^(2))`

B

`v_(0)e^(-(B_(0)^(2)l^(2)t)/(mR)`

C

`v_(0) xx e^((-2B_(0)^(2)l^(2)t)/(mR)`

D

`v_(0)e^(-(B_(0)^(2)l^(2)t)/(2mR)`

Text Solution

Verified by Experts

The correct Answer is:
B
Let at any time `t`, velocity of rod be `v`, then emf developed across its ends is `e = B_(0)vl`. Due to this induced emf, a current will establish in circuit given by, `I = (B_(0) v l)/(R )`.

For rod,
`m(d v)/(dt) = - IB_(0)l = -(B_(0)^(2)l^(2))/(R )v`
`rArr underset(v_(0))overset(v)(int)(dv)/(v) = -underset(0)overset(t)(int)(B_(0)^(2)l^(2))/(mR)dt rArr v = v_(0)e^(-(B_(0)^(2)l^(2))/(mR))`
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