Question

A long straight wire carrying current of 30 ampere is placed in an external uniform magnetic field of induction 4 into 10 to the power minus 40 the magnetic field is acting parallel to the direction of current the magnitude of the resultant magnetic induction at a point 2 cm away from the wire is

Answer 1

5 × $10^{-4}$T

Explanation:

Given that,

μ0 = 4π × 10–7T-m/A,

I = 30A,

B = 4.0 × 10–4T Parallel to current

Vector B because of wore at point 2cm;

⇒ μ0I/2πr = (4π × $10^{-7}$ × 30)/2π × 0.02

∵ Net field = $\sqrt{(3 * 10^{-4})^{2} + ( 4 * 10^{-4} )^{2} }$

= 5 × $10^{-4}$T

Learn more: find the magnitude

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