Question

A long straight wire of radius `a` carries a steady current `i`. The current is uniformly distributed across its cross section. The ratio of the magnetis field at `(a)//(2) and (2a)` is
A. `( 1//2)`
B. `(1//4)`
C. `4`
D. `1`

Answer 1

Correct Answer - D
Here , current is uniformly distributed across the cross- section of the wire, therfore, current enclosed in the amperean path formed at a distance`r_(1) ( = (a)/(2))`
` =(( pi r_(1)^(2))/( pi a^(2))xxI`, where `I` is total current
:. Magnetic field at `P_(1)` is
`B_(1) = (mu_(0) xx current enclosed)/( path) `
rArr `B_(1) =(mu_(0)xx(( pi r_(1)^(2))/( pi a^(2))) xxI)/( 2 pi r_(1)) = ( mu_(0) xx Ir_(1))/( 2 pi a^(2))`
Now, magnetic field at point `P_(2)`,
`B_(2) = (mu_(0))/( 2 pi) . (I)/( (2a)) = ( mu_(0)I)/(4 pi a)`
:. Required ratio = (B_(1))/(B_(2)) = (mu_(0)I r^(1))/( 2pia^(2))xx( 4pia)/( mu_(0)I)`
`= (2r_(1)) /(a) = (2xx (a)/(2))/(a) = 1`.