A rectangular loop with a sliding connector of Iength 10 cm is situated in a uniform magnetic field perpendicular to plane of loop. The magnetic induction is 0.1 T and resistance of connector (R) is 1 ohm. The sides AB and CD have resistances 2 ohm and 3 ohm respectively. Find the current in the connector during its motion with constant velocity one metre/sec.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1}{110} A$

$\frac{1}{220} A$

$\frac{1}{440} A$

$\frac{1}{55} A$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

When the connector moves through a distance x, then area covered = I x
flux linked = B l x = $ϕ$
Induced e.m.f., $e = \frac{dΦ}{dt} = Bl \frac{dx}{dt} = Blv$
This acts as a cell of e.m.f. B I v and internal resistance fl. Effective resistance of the circuit
$= \frac{R_{1} R_{2}}{(R_{1} + R_{2})}$
Total resistance of the circuit
$= R + \{\frac{R_{1} R_{2}}{(R_{1} + R_{2})}\}$
$∴$ current $i = \frac{Blv}{[R + \{\frac{R_{1} R_{2}}{(R_{1} + R_{2})}\}]}$
$= \frac{0 \cdot 1 \times (\frac{1}{10}) \times 1}{[1 + \{\frac{(2 \times 3)}{(2 + 3)}\}]} = \frac{1}{220} A$