Electric Field Due to an Arc at the Centre

Q.

Find the electric field at a given point ‘O’ due to a semi-circular charged ring having linear charge density $\lambda$.

1. $\frac{\lambda }{4\pi {\epsilon }_{0}r}$

2. $\frac{\lambda }{2{\epsilon }_{0}r}$

3. $\frac{\lambda }{2\pi {\epsilon }_{0}r}$

4. $\frac{\lambda }{4\pi {\epsilon }_0{r}^2}$

Q. Figure shows current loop having two circular arcs joined by two radial lines. Find the magnetic field $B$ at the center $\text{O}$. [Take $R_1=5\text{cm},R_2=10\text{cm},I=\frac{2}{\pi }\text{A}$ and $\varphi =\frac{\pi }{4}\text{rad}$]

1. $5\mu \text{T}$
2. $0.5\mu \text{T}$
3. Zero
4. $0.05\mu \text{T}$

Q.

A circular arc has a radius of 0.1m. The angle it subtends at the Centreis $\frac{\pi }{6}$ . If the charge it carries is 10C. what is the linear charge density of the wire?

1. 500 c/m

2. 50 c/m

3. 192.3 c/m

4. 215.2 c/m

Q. A thin glass rod is bent into a semicircle of radius r. A charge +Q is uniformly distributed along the upper half and a charge –Q is uniformly distributed along the lower half, as shown in the figure. The electric field E at, P (the centre of the semicircle, is)

1. $\frac{Q}{{\pi }^2{\epsilon }_0{r}^2}$
2. $\frac{2Q}{{\pi }^2{\epsilon }_0{r}^2}$
3. $\frac{4Q}{{\pi }^2{\epsilon }_0{r}^2}$
4. $\frac{Q}{4{\pi }^2{\epsilon }_0{r}^2}$