A long, vertical straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of (`4.0 xx (10^-4)` T exists from south to north. Find magnetic field at a point 2.0 away from the wire.

Explanation

Given uniform magnetic field `B = 4.0 xx10^(-4) T`
`. Magnetic field at 2.0 cm away from wire ` B=(mu_0i)/(2pid)` `
`. `=(2xx10^(-7) xx 30 )/(2xx10^(-2)`
`. `= 3xx 106(-4) T`
`. perpendicular, to B So resultant magnetic field `=sqrtB^2 + B^2`
`. `= sqrt(4xx10^(-4)^2 + (3xx 10^(-4))^2`
`. `= 5xx10^(-4)T`