A copper wire of diameter 1.6mm carries a current of 20A. Find the maximum magnitude of the magnetic field B due to this current.

 By using the Biot-Savart law

B = μ0NI/2R

We can find a formula for current carrying straight wire to find the maximum magnetic strength, that is

B = μ0i/2πr……(i)

Now, according to question,

d = 1.6mm = 1.6×10−3m

Therefore, r = 1.62mm = 0.8×10−3m

Now by putting this value in the equation (i) , we get

B = μ0×20/2×π×0.8×10−3……(ii)

Now put value of constant μ0=4π×10−7N/A2 , equation (ii) becomes

B = 4π × 10−7× 20/2 × π× 0.8 × 10−3……(iii)

∴ B = 0.005T