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Question: A current of 1 ampere is passed through a straight wire of length \(2.0m\). What will be the magneti...

A current of 1 ampere is passed through a straight wire of length 2.0m2.0m. What will be the magnetic field at a point in air at a distance of 0.3m0.3m from either end of the wire and lying on the axis of wire ?
(A) μ02π\dfrac{{{\mu _0}}}{{2\pi }}
(B) μ04π\dfrac{{{\mu _0}}}{{4\pi }}
(C) μ08π\dfrac{{{\mu _0}}}{{8\pi }}
(D) zero

Explanation

Solution

In order to solve this problem first draw a diagram in which point P is shown perfectly with wire.
After that use the formula of the magnetic field due to wire derived with the help of Biot – Savart law which is :
dB=μ04πI(d×r)r3d\vec B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{I(d\vec \ell \times \vec r)}}{{{r^3}}}
Now put the angle between d\vec d\ell and r\vec r zero because the P point is situated at the axis of the wire.
Finally we get a magnetic field at point P.

Complete step by step answer:
We know that the magnetic field due to wire is given as
dB=μ04πI(d×r)r3d\vec B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{I(d\vec \ell \times \vec r)}}{{{r^3}}} …..(1)

Here direction of d\vec d\ell and r\vec r is same i.e., angle between then is zero, so, the cross product of them is zero i.e., d×r=drsinθ\vec d\ell \times \vec r = |\vec d\ell ||\vec r|\sin \theta
=drsin0= |\vec d\ell ||\vec r|\sin 0
d×r=0\vec d\ell \times \vec r = 0 …..(2)
So, from equation 1 & 2, we get
dB=0dB = 0
B=0\therefore B = 0
Hence, the magnetic field at point P i.e., at the axis of the wire is zero.
So, option D is the correct answer zero.

Note: Here students may get confused between methods of calculating magnetic fields due to wire.
There are 2 laws :
(1) Biot – Savart law
(2) Ampere’s Circuital law
-The second point at which student may get confused is that where the point is situated i.e.
-If point is situated at the axial line then magnetic field B=0B = 0 because θ=0\theta = {0^ \circ }
-If point is not at the axis of wire then magnetic field due to wire at any point is given as
BP=μ0I2πr{B_P} = \dfrac{{{\mu _0}I}}{{2\pi r}} because θ=90\theta = {90^ \circ }