Question

A long, straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of 4.0 × 10⁻⁴ T parallel to the current. Find the magnitude of the resultant magnetic field at a point 2.0 cm away from the wire.

Answer 1

Given:
Uniform magnetic field, B₀ = 4.0 × 10⁻⁴ T 
Magnitude of current, I = 30 A
Separation of the point from the wire, d = 0.02 m
Thus, the magnetic field due to current in the wire is given by

\[B   =   \frac{\mu_0 I}{2\pi d}\]

\[=   \frac{2 \times {10}^{- 7} \times 30}{0 . 02}\] 

\[ =   3 \times  {10}^{- 4}   T\]

B₀ is perpendicular to B (as shown in the figure).
∴ Resultant magnetic field

\[B_{net}  = \sqrt{B^2 + {B_0}^2}\] 

\[             = \sqrt{(4 \times {10}^{- 4} )^2 + (3 \times {10}^{- 4} )^2}\] 

\[             = 5 \times  {10}^{- 4}   T\]