Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Need clarity, kindly explain! Two long straight parallel wires, carrying (adjustable) currents I1 and I2, are kept at a distance d apart. If the force ‘F’ between the two wires is taken as ‘positive’ when the wires repel each

Two long straight parallel wires, carrying (adjustable) currents I1 and I2, are kept at a distance d apart. If the force ‘F’ between the two wires is taken as ‘positive’ when the wires repel each other and ‘negative’ when the wires attract each other, the graph showing the dependence of ‘F’, on the product  I1I2,  would be :

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 

Answers (1)

As we learnt in

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 

 

 I1 I2 = Positive

Here the force will be attractive and hence it will be taken as negative.

In the second case  I1 I2 = Negative

Here the force will be repulsive and hence it will be taken as positive.

 


Option 1)

Incorrect

Option 2)

Incorrect

Option 3)

Incorrect

Option 4)

Correct

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks
anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?
anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question