Q. 305.0( 2 Votes )

A long, straight wire is fixed horizontally and carries a current of 50.0 A. A second wire having linear mass density 1.0 × 10–4 kg m–1 is placed parallel to and directly above this wire at a separation of 5.0 mm. What current should this second wire carry such that the magnetic repulsion can balance its weight?

. A square loop PQRS carrying a current of 6.0 A is placed near a long wire carrying 10 A as shown in the figure.

(a) Show that the magnetic force acting on the part PQ is equal and opposite to that on the part Rs.

(b) Find the magnetic force on the square loop.



Answer :

The current in the coil is 10A


There will be induction of magnetic force due to two current carrying wire placed in each other’s vicinity.


The force is given by



Where l is the length of the wire and i1 i2 are the currents flowing through the wires.


To balance the weight, the gravitational force should be equal to the magnetic force that is exerted.


Hence



Where λ is the linear mass density of the wire.


Hence, equating these forces, we get,


=



Substituting the numerical values, we get





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