# Consider th

Given:

Mass of PQ = m

Resistance of PQ = r

Length of PQ between the two rails = l

Magnetic field = B

Resistance connected to the rails = R

Velocity with which PQ is pushed towards right at t=0 = v0

Formula used:

(a) By Ohm’s law, E = IR’, where E = emf, I = current, R’ = total resistance.

Hence, current … (i)

Now, emf induced due to the moving road in the magnetic field … (ii), where B = magnetic field, l = length of rod, v= velocity of rod

Also, total resistance … (iii), where r = resistance of PQ, R = resistance attached to the rails.

Hence, substituting the values of E and R’ from (ii) and (iii) in (i), we get

Therefore, current in the loop when the speed of the wire PQ is . (Ans)

(b) Now, magnetic force on a current carrying wire … (i), where I = current, l = length of wire, B = magnetic field.

From the previous part, the value of current at an instant when velocity = v is … (ii)

Therefore, from (i) and (ii), magnetic force … (iii)

According to Newton’s second law of motion, … (iv), where F = force, m = mass, a = acceleration.

Hence, equating (iii) and (iv):

Therefore, acceleration of the wire at this instant = (Ans)

(c) Velocity v’ can be expressed as … (i), where v0 = initial velocity, a = acceleration, t = time. We put a negative sign before at since the force is opposite to velocity, and it

Now, from the previous part, we can write acceleration … (ii)

Hence, from (i) and (ii), we can write

But, distance travelled x = vt, where v = velocity, t = time.

Therefore, velocity v as a function of x is (Ans)

(d) We know that,, where a = acceleration,

v = velocity, x = distance, t = time.

Now, from the part (b), acceleration as a function of time

Now, the wire can travel maximum distance when its velocity is v0.

Hence, integrating on both sides, we get

Therefore, maximum distance travelled by the wire

(Ans)

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