Current density in a cylindrical wire of radius {R} varies with radial distance as \beta(r+r_{0})^{2}. The current through the section of the wire shown in the figure is.

Correct Answer is : πβ[R412+r02R26+2r0R39]

\pi \beta[ { {R}^{4}}/{12}+ { {r}_{0}^{2} {R}^{2}}/{6}+ {2 {r}_{0} {R}^{3}}/{9}]

\pi \beta[ { {R}^{4}}/{6}+ { {r}_{\circ}^{2} {R}^{2}}/{12}+ { {r}_{0} {R}^{3}}/{9}]

\pi \beta[\frac{R^{4}}{12}+ {r_{0}^{2} R^{2}}/{12}+ {r_{0} R^{3}}/{9}]

\pi \beta[ { {R}^{4}}/{8}+ { {r}_{0}^{2} {R}^{2}}/{6}+ { {r}_{0} {R}^{3}}/{12}]

AP EAPCET 04-Jul-2022 Shift 1 Paper

Section: Physics

Question : 107 of 160

Marks: +1, -0

Current density in a cylindrical wire of radius R varies with radial distance as β(r+r0)2. The current through the section of the wire shown in the figure is.

πβ[‌
R4
12
+‌
r02R2
6
+‌
2r0R3
9
]
πβ[‌
R4
6
+‌
r∘2R2
12
+‌
r0R3
9
]
πβ[‌
R4
12
+‌
r02R2
12
+‌
r0R3
9
]
πβ[‌
R4
8
+‌
r02R2
6
+‌
r0R3
12
]

Validate

Go to Question:

Prev Question

Next Question