A metal wire PQ slides on parallel metallic rails having separation 0.25 m, each having negligible resistance. There is a 2 Ω resistor and 10 V battery as shown in the figure. There is a uniform magne

Question

A metal wire PQ slides on parallel metallic rails having separation $0.25$ m, each having negligible resistance. There is a $2 Ω$ resistor and $10$ V battery as shown in the figure. There is a uniform magnetic field directed into the plane of the paper of magnitude $0.5$ T. A force of $0.5$ N to the left is required to keep the wire PQ moving with constant speed to the right. With what speed (in m/s) is the wire PQ moving? (Neglect self inductance of the loop.)

Answer:

16

Solution:

Induced emf:
$B ℓ v = 0.125 v$
Current:
$I = \frac{10 - 0.125 v}{2}$
Force:
$B I ℓ = 0.5 (\frac{10 - 0.125 v}{2}) 0.25 = 0.5$ N (given)
Solving:
$v = 16$ m/s

TestHub · Accepted Answer

Induced emf: B ℓ v = 0.125 v Current: I = 2 10 − 0.125 v ​ Force: B I ℓ = 0.5 ( 2 10 − 0.125 v ​ ) 0.25 = 0.5 N (given) Solving: v = 16 m/s

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JEE Main 2027 (Class 12+)