Question

A long straight wire (radius = 3.0 mm) carries a constant current distributed uniformly over a crea section
perpendicular to the axis of the wire. If the current density is 100 A/m2. The magnitudes of we magnetic
field at (a) 2.0 mm from the axis of the wire and (b) 4.0 mm from the axis of the wire :-
a) 2π x 10^-8 T , 9π/2 x 10^-8 T
b) 4π x10^-8 T, π/2 x10^-8 T
c) 4π x 10^-8 T, 9π/2 x10^-8 T
d) π x 10^-4 T, 9π/2 x 10^-8 T​

Answer 1

Given that,

Radius = 3.0 mm

Current density = 100 A/m²

(a). 2.0 mm from the axis of the wire

(b). 4.0 mm from the axis of the wire

Current in the wire :

Current is defined as,

$I=JA$

$I=J(\pi r^2)$

Where, I = current

J = current density

r = radius

(a). We need to calculate the magnetic field

Using formula of magnetic field

$\int{B. dL}=\mu_{0} I$

$B(2\pi r)=\mu_{0}\times J(\pi r^2)$

$B=\dfrac{\mu_{0}J(\pi r^2)}{2\pi r}$

$B=\dfrac{\mu_{0}Jr}{2}$

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times100\times2\times10^{-3}}{2}$

$B= 4\pi\times10^{-8}\ T$

(b). We need to calculate the magnetic field

Using formula of magnetic field

$\int{B. dL}=\mu_{0} I$

$B(2\pi r)=\mu_{0}\times J(\pi R^2)$

$B=\dfrac{\mu_{0}J(R^2)}{2r}$

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times100\times(3\times10^{-3})^2}{2\times4.0\times10^{-3}}$

$B=\dfrac{9\pi}{2}\times10^{-8}\ T$

Hence, The magnetic fields are $4\pi\times10^{-8}\ T$ and $\dfrac{9\pi}{2}\times10^{-8}\ T$

(c) is correct option