BITSAT 2019 Slot 2 Solved Paper

Given, current through the square loop is 2A.
The given situation is shown in the figure. Magnetic field due to wire AB at the centre O is
BAB=

µ0

4π

.

I

r

[sin‌45°+sin‌45°]
=10−7×

2

2×10−2

[

1

√2

+

1

√2

]
=√2×10−5T [Downward to the plane of square loop]
Since, magnetic field due to side AB,BC,CD and DA is same in magnitude and direction. Hence, net magnetic field at O
B=4×BAB=4√2×10−5T
Wheatstone bridge in balanced condition maintains

P

Q

=

R

S

i.e. no current flows through the galvanometer (i.e. ig=0 )