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  • (i) Depict magnetic field lines due to two straight, long. parallel conductors carrying steady currents and <img alt="I_{2}" src="https://entrancecorner.codecogs.com/

(i) Depict magnetic field lines due to two straight, long. parallel conductors carrying steady currents I_{1}and I_{2} in the same direction.

(ii) Write the expression for the magnetic field produced by one of the conductor over the other. Deduce an expression for the force per unit length.
(iii) Determine the direction of this force.

(iv) In figure given below, wire PQ is fixed while the square loop ABCD is free to move under the influence ol currents flowing in them. State with reason, in which direc tion does the loop be gin to move or rotate?

 

 

 

 
 
 
 
 

Answers (1)

(i)

(ii)

B_{1}=\frac{\mu_{o}I_{1}}{2\pi d}

B_{2}=\frac{\mu_{o}I_{2}}{2\pi d}

F_{12}=I_{2}B_{1}L

F_{21}=I_{1}B_{2}L

F = F_{12}=F_{21}=\frac{\mu_{o}II_{2}L}{2\pi d}

\frac{f}{L}=\frac{\mu_{o}I_{1}I_{2}}{2\pi d}

The force is attractive.

(iii) \rightarrow The loop will move towards PQ

\rightarrow The force due to AB and CD cancel each other.

\rightarrowThe force between PQ and AD are attractive

\rightarrowThe force between PQ and BC are repulsive, but the magnitude of force due to BC is less than that of AB. So block move towards PQ.

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