Question

Question 4.2: A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-169

Answer 1

we know formula
magnetic field due to straight line wire is given by $\mathbb{B}=\frac{\mu_0}{4\pi}\frac{2I}{r}$

here B is the magnetic field,
$\mu_0$ is the permeability of medium in vaccum,
r is the separation between wire and observation point.

so, $B = 1\times10^{-7}\times\frac{2\times35}{20\times10^{-2}}\\\\=3.5\times10^{-5}T$

hence, B = 3.5 × 10^-5 T

Answer 2

Current in the wire, I = 35 A

Distance of a point from the wire, r = 20 cm = 0.2 m

Magnitude of the magnetic field at this point is given as:

B  

Where,

μ0= Permeability of free space = 4π × 10–7 T m A–1

Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 × 10–5 T.