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 Three straight parallel current carrying  conductors are shown in the figure. The force experienced by the middle conductor of length 25 cm is :

  • Option 1)

       3 x 10-4  N  toward right

  • Option 2)

       6 x 10-4  N  toward left

  • Option 3)

       9 x 10-4  N  toward left

  • Option 4)

    Zero

 

Answers (1)

best_answer

As we have learned

Force between two parallel current carrying conductors -

F=\frac{\mu }{4\pi } \frac{2I_{1 I_{2}}}{a} l

\frac{F}{l}=\frac{\mu o}{4\pi } \frac{2I_{1 I_{2}}}{a}

- wherein

I1 and I2 current carrying two parallel wires 

a-seperation between two wires 

 

 

Force due to wire I 

 F _1= \frac{\mu _0 I_1 I_2 }{2\pi r_1 }l= \frac{2 \times 10^{-7}\times 30\times 10}{3\times 10^{-2}}\times 25\times 10^{-2}= 5\times 10^{-4} towards right 

Force due to wire two 

F _2= \frac{\mu _0 I I_2 }{2\pi r_2 }l= \frac{2 \times 10^{-7}\times 20\times 10}{5\times 10^{-2}}\times 25\times 10^{-2}= 2\times 10^{-4}  towards left 

Net force = 3 \times 10^{-4} \: \: towards \: \: right

 

 

 

 

 

 

 


Option 1)

   3 x 10-4  N  toward right

Option 2)

   6 x 10-4  N  toward left

Option 3)

   9 x 10-4  N  toward left

Option 4)

Zero

Posted by

Avinash

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