Question

At what distance from a long straight wire carrying current of 12A will the magnetic field be equal to $3\times 10^{-5} Wb/m^2$?
A. 8 cm
B. 12 cm
C. 18 cm
D. 24 cm

Answer 1

Hint: Magnetic field is the space around the magnet where the effect of magnet can be felt by another magnet or iron piece. Magnetic fields can also be produced by a moving charge whose intensity can be determined by the velocity and magnitude of charge. The S.I unit of magnetic field is Tesla (T) whereas the C.G.S unit is Gauss (G).
Formula used:
$B = \dfrac{\mu_{\circ}i}{4\pi d}(sin \phi_1 + sin \phi_2)$

Complete answer:
We will proceed by understanding the formula$B = \dfrac{\mu_{\circ}i}{4\pi d}(sin \phi_1 + sin \phi_2)$. Here B is the net magnetic field due to current ‘i’ which is at a distance ‘d’. Note that the distance must be the shortest distance between the wire and point. The magnetic field due to a current carrying wire at some distance is given by Biot-Savart's law. Using this law, one can find the magnetic field values due to any number of wires. For using this law, one must know the shortest distance of the point to the wire which can be obtained by dropping a perpendicular from the point onto the wire. For a long wire, both $\phi_1 = \phi_2 = 90^{\circ}$.
Thus, the formula becomes $B = \dfrac{\mu_{\circ}i}{2\pi d}$.
Now, given: B = $3\times 10^{-5} Wb/m^2$, $i=12A$
Thus, putting the values in the equation, we get;
$3\times 10^{-5} = \dfrac{4\pi\times 10^{-7}(12)}{2\pi d}$
$d = 8\times 10^{-2}m = 8cm$

So, the correct answer is “Option A”.

Note:
The term $\mu_{\circ}$, which is the permeability of free space has value $4\pi \times 10^{-7}m \ kgs^{-2}A^{-2} = 4\pi \times 10^{-7} H/m$, where H stands for henry, the unit of inductance. Note, the expression $B = \dfrac{\mu_{\circ}i}{2\pi d}$ could be used if the wire is long enough or the distance of the point is very small.