153179
A coil having 100 turns is wound tightly in the form of a spiral with inner and outer radii $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$, respectively. When a current $1 \mathrm{~A}$ passes through the coil, the magnetic field at the centre of the coil is
C Magnetic field at point $\mathrm{O}=$ Magnetic field at $\mathrm{O}$ due to straight wire + Magnetic field of wire of loop $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}+\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(\frac{3 \pi}{2}\right)$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}+\frac{3}{8} \frac{\mu_{0} \mathrm{I}}{\mathrm{R}}$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(1+\frac{3 \pi}{2}\right)$
Karnataka CET-2020
Moving Charges & Magnetism
153176
The magnetic field at point $P$ of given figure due to carrying of current $I$ by a conductor of radius $R$, is
B The magnetic field at point $P$ distance $r$ of given figure due to carrying current $\mathrm{I}$ by a conductor of Radius $\mathrm{R}$ is given by $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{T}$
TS- EAMCET-09.09.2020
Moving Charges & Magnetism
153181
A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ of the loop is
A Magnetic field due to $i_{1}=\frac{\mu_{0} i_{1} \theta_{1}}{2 R 2 \pi}$ $\text { Magnetic field due to } i_{2}=\frac{\mu_{0} i_{2} \theta_{2}}{2 R 2 \pi}$ For parallel combination- $\frac{\mathrm{i}_{1}}{\mathrm{i}_{2}}=\frac{\rho \mathrm{l}_{2}}{\mathrm{~A}} \times \frac{\mathrm{A}}{\rho \mathrm{l}_{1}}=\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}}$ $\frac{\mathrm{i}_{1}}{\mathrm{i}_{2}}=\frac{\frac{1}{4}(2 \pi \mathrm{R})}{\frac{3}{4}(2 \pi \mathrm{R})}$ $\mathrm{i}_{1}=\frac{\mathrm{i}_{2}}{3}, \quad \mathrm{i}_{2}=3 \mathrm{i}_{1}$ Net magnetic field- $=\frac{\mu_{0} \mathrm{i}_{1}}{2 \mathrm{R}}\left(\frac{\theta_{1}}{2 \pi}\right)-\frac{\mu_{0} \mathrm{i}_{2}}{2 \mathrm{R}}\left(\frac{\theta_{2}}{2 \pi}\right)$ $=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \pi}{2 \times 2 \pi}\right)-\frac{\mu_{0} \mathrm{i}_{2}}{2 \mathrm{R}}\left(\frac{\pi}{2 \pi}\right) \times \frac{1}{2}$ $=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \mathrm{i}_{1}}{4}-\frac{\mathrm{i}_{2}}{4}\right)=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \mathrm{i}_{1}}{4}-\frac{3 \mathrm{i}_{1}}{4}\right)=0$
153179
A coil having 100 turns is wound tightly in the form of a spiral with inner and outer radii $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$, respectively. When a current $1 \mathrm{~A}$ passes through the coil, the magnetic field at the centre of the coil is
C Magnetic field at point $\mathrm{O}=$ Magnetic field at $\mathrm{O}$ due to straight wire + Magnetic field of wire of loop $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}+\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(\frac{3 \pi}{2}\right)$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}+\frac{3}{8} \frac{\mu_{0} \mathrm{I}}{\mathrm{R}}$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(1+\frac{3 \pi}{2}\right)$
Karnataka CET-2020
Moving Charges & Magnetism
153176
The magnetic field at point $P$ of given figure due to carrying of current $I$ by a conductor of radius $R$, is
B The magnetic field at point $P$ distance $r$ of given figure due to carrying current $\mathrm{I}$ by a conductor of Radius $\mathrm{R}$ is given by $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{T}$
TS- EAMCET-09.09.2020
Moving Charges & Magnetism
153181
A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ of the loop is
A Magnetic field due to $i_{1}=\frac{\mu_{0} i_{1} \theta_{1}}{2 R 2 \pi}$ $\text { Magnetic field due to } i_{2}=\frac{\mu_{0} i_{2} \theta_{2}}{2 R 2 \pi}$ For parallel combination- $\frac{\mathrm{i}_{1}}{\mathrm{i}_{2}}=\frac{\rho \mathrm{l}_{2}}{\mathrm{~A}} \times \frac{\mathrm{A}}{\rho \mathrm{l}_{1}}=\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}}$ $\frac{\mathrm{i}_{1}}{\mathrm{i}_{2}}=\frac{\frac{1}{4}(2 \pi \mathrm{R})}{\frac{3}{4}(2 \pi \mathrm{R})}$ $\mathrm{i}_{1}=\frac{\mathrm{i}_{2}}{3}, \quad \mathrm{i}_{2}=3 \mathrm{i}_{1}$ Net magnetic field- $=\frac{\mu_{0} \mathrm{i}_{1}}{2 \mathrm{R}}\left(\frac{\theta_{1}}{2 \pi}\right)-\frac{\mu_{0} \mathrm{i}_{2}}{2 \mathrm{R}}\left(\frac{\theta_{2}}{2 \pi}\right)$ $=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \pi}{2 \times 2 \pi}\right)-\frac{\mu_{0} \mathrm{i}_{2}}{2 \mathrm{R}}\left(\frac{\pi}{2 \pi}\right) \times \frac{1}{2}$ $=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \mathrm{i}_{1}}{4}-\frac{\mathrm{i}_{2}}{4}\right)=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \mathrm{i}_{1}}{4}-\frac{3 \mathrm{i}_{1}}{4}\right)=0$
153179
A coil having 100 turns is wound tightly in the form of a spiral with inner and outer radii $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$, respectively. When a current $1 \mathrm{~A}$ passes through the coil, the magnetic field at the centre of the coil is
C Magnetic field at point $\mathrm{O}=$ Magnetic field at $\mathrm{O}$ due to straight wire + Magnetic field of wire of loop $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}+\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(\frac{3 \pi}{2}\right)$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}+\frac{3}{8} \frac{\mu_{0} \mathrm{I}}{\mathrm{R}}$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(1+\frac{3 \pi}{2}\right)$
Karnataka CET-2020
Moving Charges & Magnetism
153176
The magnetic field at point $P$ of given figure due to carrying of current $I$ by a conductor of radius $R$, is
B The magnetic field at point $P$ distance $r$ of given figure due to carrying current $\mathrm{I}$ by a conductor of Radius $\mathrm{R}$ is given by $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{T}$
TS- EAMCET-09.09.2020
Moving Charges & Magnetism
153181
A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ of the loop is
A Magnetic field due to $i_{1}=\frac{\mu_{0} i_{1} \theta_{1}}{2 R 2 \pi}$ $\text { Magnetic field due to } i_{2}=\frac{\mu_{0} i_{2} \theta_{2}}{2 R 2 \pi}$ For parallel combination- $\frac{\mathrm{i}_{1}}{\mathrm{i}_{2}}=\frac{\rho \mathrm{l}_{2}}{\mathrm{~A}} \times \frac{\mathrm{A}}{\rho \mathrm{l}_{1}}=\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}}$ $\frac{\mathrm{i}_{1}}{\mathrm{i}_{2}}=\frac{\frac{1}{4}(2 \pi \mathrm{R})}{\frac{3}{4}(2 \pi \mathrm{R})}$ $\mathrm{i}_{1}=\frac{\mathrm{i}_{2}}{3}, \quad \mathrm{i}_{2}=3 \mathrm{i}_{1}$ Net magnetic field- $=\frac{\mu_{0} \mathrm{i}_{1}}{2 \mathrm{R}}\left(\frac{\theta_{1}}{2 \pi}\right)-\frac{\mu_{0} \mathrm{i}_{2}}{2 \mathrm{R}}\left(\frac{\theta_{2}}{2 \pi}\right)$ $=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \pi}{2 \times 2 \pi}\right)-\frac{\mu_{0} \mathrm{i}_{2}}{2 \mathrm{R}}\left(\frac{\pi}{2 \pi}\right) \times \frac{1}{2}$ $=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \mathrm{i}_{1}}{4}-\frac{\mathrm{i}_{2}}{4}\right)=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \mathrm{i}_{1}}{4}-\frac{3 \mathrm{i}_{1}}{4}\right)=0$
153179
A coil having 100 turns is wound tightly in the form of a spiral with inner and outer radii $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$, respectively. When a current $1 \mathrm{~A}$ passes through the coil, the magnetic field at the centre of the coil is
C Magnetic field at point $\mathrm{O}=$ Magnetic field at $\mathrm{O}$ due to straight wire + Magnetic field of wire of loop $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}+\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(\frac{3 \pi}{2}\right)$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}+\frac{3}{8} \frac{\mu_{0} \mathrm{I}}{\mathrm{R}}$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(1+\frac{3 \pi}{2}\right)$
Karnataka CET-2020
Moving Charges & Magnetism
153176
The magnetic field at point $P$ of given figure due to carrying of current $I$ by a conductor of radius $R$, is
B The magnetic field at point $P$ distance $r$ of given figure due to carrying current $\mathrm{I}$ by a conductor of Radius $\mathrm{R}$ is given by $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{T}$
TS- EAMCET-09.09.2020
Moving Charges & Magnetism
153181
A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ of the loop is
A Magnetic field due to $i_{1}=\frac{\mu_{0} i_{1} \theta_{1}}{2 R 2 \pi}$ $\text { Magnetic field due to } i_{2}=\frac{\mu_{0} i_{2} \theta_{2}}{2 R 2 \pi}$ For parallel combination- $\frac{\mathrm{i}_{1}}{\mathrm{i}_{2}}=\frac{\rho \mathrm{l}_{2}}{\mathrm{~A}} \times \frac{\mathrm{A}}{\rho \mathrm{l}_{1}}=\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}}$ $\frac{\mathrm{i}_{1}}{\mathrm{i}_{2}}=\frac{\frac{1}{4}(2 \pi \mathrm{R})}{\frac{3}{4}(2 \pi \mathrm{R})}$ $\mathrm{i}_{1}=\frac{\mathrm{i}_{2}}{3}, \quad \mathrm{i}_{2}=3 \mathrm{i}_{1}$ Net magnetic field- $=\frac{\mu_{0} \mathrm{i}_{1}}{2 \mathrm{R}}\left(\frac{\theta_{1}}{2 \pi}\right)-\frac{\mu_{0} \mathrm{i}_{2}}{2 \mathrm{R}}\left(\frac{\theta_{2}}{2 \pi}\right)$ $=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \pi}{2 \times 2 \pi}\right)-\frac{\mu_{0} \mathrm{i}_{2}}{2 \mathrm{R}}\left(\frac{\pi}{2 \pi}\right) \times \frac{1}{2}$ $=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \mathrm{i}_{1}}{4}-\frac{\mathrm{i}_{2}}{4}\right)=\frac{\mu_{0}}{2 \mathrm{R}}\left(\frac{3 \mathrm{i}_{1}}{4}-\frac{3 \mathrm{i}_{1}}{4}\right)=0$
