The figure shows a conducting loop ABCDA placed in a uniform magn… - Vidyadip

MCQ

The figure shows a conducting loop $ABCDA$ placed in a uniform magnetic field perpendicular to its plane. The part $ABC$ is the $(3/4)^{th}$ portion of the square of side length $l$ . The part $ADC$ is a circular arc of radius $R$ . The points $A$ and $C$ are connected to a battery which supply a current $I$ to the circuit. The magnetic force on the loop due to the field $B$ is

A
Zero
✓
$BI l$
C
$2BIR$
D
$\frac{{BI/R}}{{l + R}}$

✓

Answer

Correct option: B.

$BI l$

b
Introducing two equal and opposite current $I_{1}$ and $a l s o I_{2}$ between $A \& C$

Force on $ABCA$ closed loop zero

Force on $ADCA$ closed loop zero

Force on extra $I_{1} \& I_{2}$

$F=\left(I_{1}+L_{2}\right) l B=I / B$

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