# A rectangular loop has a sliding connector PQ of length $\dpi{100} l$ and resistance $\dpi{100} R\: \Omega$ and it is moving with a speed $\dpi{100} \upsilon$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $\dpi{100} I_{1},I_{2}\: and \: I_{3}$ are Option 1) $I_{1}=I_{2}= \frac{Bl\upsilon }{6R},I= \frac{Bl\upsilon }{3R}$ Option 2) $I_{1}=-I_{2}= \frac{Bl\upsilon }{R},I= \frac{2Bl\upsilon }{R}$ Option 3) $I_{1}=I_{2}= \frac{Bl\upsilon }{3R},I= \frac{2Bl\upsilon }{3R}$ Option 4) $I_{1}=I_{2}= I= \frac{Bl\upsilon }{R},$

As we learnt in

Motional EMF -

$\varepsilon = Blv$

- wherein

$B\rightarrow$ magnetic field

$l\rightarrow$ length

$v\rightarrow$ velocity perpendicular to uniform magnetic field.

$I={}\frac{\xi{}3R}{2}=\frac{2\xi }{3R}=\frac{2Bl\vartheta }{3R}$

$I_{1}=I_{2}=\frac{I}{2}=\frac{Bl\vartheta }{3R}$

Option 1)

$I_{1}=I_{2}= \frac{Bl\upsilon }{6R},I= \frac{Bl\upsilon }{3R}$

This is incorrect option

Option 2)

$I_{1}=-I_{2}= \frac{Bl\upsilon }{R},I= \frac{2Bl\upsilon }{R}$

This incorrect option

Option 3)

$I_{1}=I_{2}= \frac{Bl\upsilon }{3R},I= \frac{2Bl\upsilon }{3R}$

This is correct option

Option 4)

$I_{1}=I_{2}= I= \frac{Bl\upsilon }{R},$

This is incorrect option

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout JEE Main April 2022 (Subscription)

Knockout JEE Main April 2022 Subscription.

₹ 5499/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-