Question

A straight vertical wire carries a current .At a point 5cm due north of it the magnetic induction is found to be 20 utdue east.The magnetic induction at a point 10cm east of it will be

Answer 1

Given : A straight vertical wire carries a current. At a point 5cm due north of it the magnetic induction is found to be 20 μT due east.

To find : The magnetic induction at a point 10 cm east of it will be...

solution : magnitude of magnetic field due to straight wire at r distance is given by, B = μ₀i/2πr

here it is clear that B ∝ 1/r

i.e., B₁/B₂ = r₂/r₁

⇒20μT/B₂ = (10cm)/(5cm)

⇒B₂ = 10μT

Therefore magnitude of magnetic induction is 10μT.

now consider , North = j , south = - j ,

East = i , West = - i ,

vertically upward = k,

vertically downward = - k

case 1 : direction of position of observation point , r = j

direction of B = i

let direction of infinitesimal length, dl = x

here, direction of B = direction of dl × r

⇒i = x × j

i.e., it is clear that if x = -k , then j × -k = i

it means direction of infinitesimal length is vertically downward direction.

now case 2 : let direction of magnetic field = y

direction of point of observation = i

and direction of dl = -k

so, y = -k × i = - j ( south)

Therefore the direction of magnetic field is south.

Therefore the magnetic induction at a point 10 cm east of it will be 10μT south.

Answer 2

Answer:

the answer will be 10 utdue