Rankers Physics
Topic: Magnetic Effects of Current
Subtopic: Ampere's Circuital Law

A hallow cylindrical wire carries current I, having inner & outer radius R & 2R respectively. Magnetic field at a point which is 5R/4 distance away from the wire :
\[\frac{5\mu_{0}I}{18\pi R}\]
\[\frac{\mu_{0}I}{36\pi R}\]
\[\frac{5\mu_{0}I}{36\pi R}\]
\[\frac{3}{40}\frac{\mu_{0}I}{\pi R}\]

Solution:

To solve for the magnetic field at a distance of 5R4\frac{5R}{4} from the axis of a hollow cylindrical wire carrying current II, with inner radius RR and outer radius 2R2R, we use Ampère's Law. The key steps are:

1. Magnetic Field Inside a Hollow Cylinder

For R<r<2RR < r < 2R (points within the shell of the cylinder):

  • Current density JJ is uniform, given by:

    J=Iπ((2R)2R2)=I3πR2.J = \frac{I}{\pi \left((2R)^2 - R^2\right)} = \frac{I}{3\pi R^2}.

  • The current enclosed within a radius rr (where R<r<2RR < r < 2R) is:

    Ienc=Jarea enclosed=Jπ(r2R2).I_{\text{enc}} = J \cdot \text{area enclosed} = J \cdot \pi (r^2 - R^2).Substituting JJ:

    Ienc=I3πR2π(r2R2)=I3R2(r2R2).I_{\text{enc}} = \frac{I}{3\pi R^2} \cdot \pi (r^2 - R^2) = \frac{I}{3R^2}(r^2 - R^2).

  • Using Ampère's Law:

    B2πr=μ0Ienc,B \cdot 2\pi r = \mu_0 I_{\text{enc}}, B=μ02πrI3R2(r2R2).B = \frac{\mu_0}{2\pi r} \cdot \frac{I}{3R^2}(r^2 - R^2).

2. Magnetic Field at r=5R4r = \frac{5R}{4}

Since 5R4\frac{5R}{4} lies within the shell (R<r<2RR < r < 2R), substitute r=5R4r = \frac{5R}{4} into the above equation:

B=μ02π5R4I3R2((5R4)2R2).B = \frac{\mu_0}{2\pi \cdot \frac{5R}{4}} \cdot \frac{I}{3R^2}\left(\left(\frac{5R}{4}\right)^2 - R^2\right).

Simplify:

  • r2=(5R4)2=25R216r^2 = \left(\frac{5R}{4}\right)^2 = \frac{25R^2}{16},
  • r2R2=25R216R2=25R21616R216=9R216r^2 - R^2 = \frac{25R^2}{16} - R^2 = \frac{25R^2}{16} - \frac{16R^2}{16} = \frac{9R^2}{16}.

Thus:

B=μ02π5R4I3R29R216.B = \frac{\mu_0}{2\pi \cdot \frac{5R}{4}} \cdot \frac{I}{3R^2} \cdot \frac{9R^2}{16}.

Simplify further:

B=μ02π45R9I48=μ0I2π36240R.B = \frac{\mu_0}{2\pi} \cdot \frac{4}{5R} \cdot \frac{9I}{48} = \frac{\mu_0 I}{2\pi} \cdot \frac{36}{240R}.

Final simplification:

B=340μ0IπR.B = \frac{3}{40} \frac{\mu_0 I}{\pi R}.

Final Answer:

B=340μ0IπR\boxed{B = \frac{3}{40} \frac{\mu_0 I}{\pi R}}

Leave a Reply

Your email address will not be published. Required fields are marked *