Question

Four long straight thin wires are held vertically at the corners A, B, C and D of a square of side ‘a’, kept on a table and carry equal current T. The wire at A carries current in upward direction whereas the current in the remaining wires flows in downward direction. The net magnetic field at the centre of the square will have the magnitude ______.

Options

• `(mu_0 I)/(pi a)` and directed along OC.

• `(mu_0 I)/(pi a sqrt 2)` and directed along OD.

• `(mu_0 I sqrt2)/(pi a)` and directed along ОВ.

• `(2 mu_0 I)/(pi a)` and directed along OA.

Answer 1

Four long straight thin wires are held vertically at the corners A, B, C and D of a square of side ‘a’, kept on a table and carry equal current T. The wire at A carries current in upward direction whereas the current in the remaining wires flows in downward direction. The net magnetic field at the centre of the square will have the magnitude `bbunderline((mu_0 I sqrt2)/(pi a) "and directed along ОВ")`.

Explanation:

Magnetic field due to a long straight current-carrying wire:

B = `(mu_0 I)/(2 pi r)`

Distance from centre to each corner:

r = `a/sqrt 2`

Thus field due to each wire:

B₀ = `(mu_0 I)/(2 pi r)`

= `(mu_0 I)/(2 pi(a/sqrt 2))`

= `(mu_0 I sqrt 2)/(2 pi a)`

Each magnetic field makes a 45° angle with the diagonals. Effective components add vectorially, giving:

B_net = 2B₀

= `2 xx (mu_0 I sqrt 2)/(2 pi a)`

= `(mu_0 I sqrt 2)/(pi a)`

From vector addition, the resultant is along the diagonal OB.