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
Rate this question :






















Two proton beams going in the same direction repel each other whereas tow wires carrying currents in the same direction attract each other. Explain.
HC Verma - Concepts of Physics Part 2Consider the situation described in the previous problem. Suppose the current i enters the loop at the point A and leaves it at the point B. Find the magnetic field at the centre of the loop.
HC Verma - Concepts of Physics Part 2Consider the situation shown in figure. The straight wire is fixed but the loop can move under magnetic force. The loop will
HC Verma - Concepts of Physics Part 2
Find the magnetic field B at the centre of a rectangular loop of length ℓ and width b, carrying a current i.
HC Verma - Concepts of Physics Part 2