Hollow cylindrical wire carries current I , having inner and outer radii R and 2R respectively. The magnetic field at a point which is (5 R/4) distance away from the wire:-

Question

Hollow cylindrical wire carries current I , having inner and outer radii R and 2R respectively. The magnetic field at a point which is (5 R/4) distance away from the wire:- (A) (5 μ 0 I/18 π R) (B) (μ 0 I/36 π R) (C) (5 μ 0 I/36 π R) (D) (3/40)(μ 0 I/π R)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-ndpj1sbimrqxqwf3.jpg" /> i textin =(Iπ ((25 R2/16) - R2)/3 π R2) i textin =(3 I/16) B=(μ0 ii n/2 π((5 R)4)) ⇒ B=(μ0(3 I / 16)/5 π R / 2)=(3 μ0 I/40 π R)

Q. Hollow cylindrical wire carries current $I$ , having inner and outer radii $R$ and $2 R$ respectively. The magnetic field at a point which is $\frac{5 R}{4}$ distance away from the wire:-

$\frac{5 μ _{0} I}{18 π R}$

$\frac{μ _{0} I}{36 π R}$

$\frac{5 μ _{0} I}{36 π R}$

$\frac{3}{40} \frac{μ _{0} I}{π R}$

$i_{in} = \frac{I π ( \frac{25 R ^{2}}{16} - R ^{2} )}{3 π R ^{2}}$
$i_{in} = \frac{3 I}{16}$
$B = \frac{μ _{0} i _{in}}{2 π ( \frac{5 R}{4} )} \Rightarrow B = \frac{μ _{0} ( 3 I /16 )}{5 π R /2} = \frac{3 μ _{0} I}{40 π R}$

Q. Hollow cylindrical wire carries current I , having inner and outer radii R and 2R respectively. The magnetic field at a point which is 45R​ distance away from the wire:-

18πR5μ0​I​

36πRμ0​I​

36πR5μ0​I​

403​πRμ0​I​