Discussion Forum

1)

 The magnitude of the force vector acting on a unit length of a thin wire carrying a current I= 8A at a point O, if the wire  is bent as shown in the figure with a radius , $R=10 \pi$ cm is 

Answer:

Option A

Explanation:

Given, current flowing  through wire, I= 8 A and radius of circular wire,

 R=$10 \pi $ cm= $10 \pi \times 10^{-2}$ m

 According to the question,

$\therefore$  Magnetic field produced by semi-circular current carrying thin wire at centre O.

 $B=\frac{\mu_{0}I}{4R}$

 Putting the given values, we get

 $B=\frac{4 \pi \times 10 ^{-7} \times 8}{4 \times 10 \pi \times 10^{-2}}$

$\Rightarrow$      B= $ 8 \times 10 ^{-6} $ T

 $\therefore$  Magnitude  of the force on thin wire per unit length.

   F= IB

 Putting the given values,

 = $8 \times 8 \times 10^{-6}$

= $64 \times 10^{-6}$ N/m=64$\mu$ N/m