Question

Find the magnitude of force acting on the conductor carrying current $I$ as shown in the diagram. The magnitude of magnetic field is $B$ and direction is into the plane of paper :

• A. $I(2L+\mu R)B$
• B. $I(2L+R)B$
• C. $I(2L+2R)B$
• D. none of these

Answer 1

The correct option is C $I(2L+2R)B$

$F_{net}=F+F+{\int }_0^{90}2I{F}_{1}cos\theta$

Force due to straight current carrying wire$=IBl$
and due to curved part, there will be two component of force
Horizontal component of right half part will be canceled by left half
Vertical component of right half and left half will be add up and resultant would be in vertical direction

$F_{net}=IlB+IlB+{\int }_0^{90}2lRd\theta cos\theta$

$=2IlB+2IRB$

Short cut $F_{net}=F+F+IB\times$ displacement length of curved part

$=IlB+IlB+IB\left(2R\right)$

$=2IlB+2IRB$.