Explanation
The number of turns in two concentric coils are {n_1}\;{\text{and}}\;{n_2} and the ratio of their radii is 2:When equal currents is passed through them in opposite direction, the magnetic field produced at the center becomes zero. The ratio {n_1}:{n_2} will be:
\textbf{Given}: Length l and magnetic moment M of steel wire
\textbf{To find}: New magnetic moment
\textbf{Solution}:
We know, pole strength is given by,
m=\dfrac{M}{l}
A steel wire is bent into semi-circular arc, so if r is radius of semicircle, then
πr=l,
Therefore, r= \dfrac{1}{\pi }
Now new magnetic moment M′=m\times 2r
=\dfrac{M}{l}\times \dfrac{2l}{\pi }
=\dfrac{2M}{\pi } =2M\pi ^{-1}
\textbf{Hence B is the correct option}
Correct Answer: A
Hint: Calculate net magnetic field direction at R and then use \vec{F} = i(\vec{dl} \times \vec{B})
Step-1: Force on R due to the current in P & Q
\therefore {\text{ Force per unit length on the wire R due to Q wire is }}
{\text{ }}{{\text{F}}_1} = \dfrac{{{\mu _{o{I_3}{I_2}}}}}{{2\pi r}} = \dfrac{{24{\mu _o}}}{{2\pi r}}{\text{ direction towards left according to Fleming's left hand rule}}
{\text{ Similarly, force per unit length on the wire R due to P wire is}}
{\text{ }}{{\text{F}}_2}{\text{ = }}\dfrac{{{\mu _o}{I_1}{I_3}}}{{2\pi \times 2r}}{\text{ = }}\dfrac{{12{\mu _o}}}{{4\pi r}}{\text{ direction towards left according to Fleming's left hand rule}}
Step-2: Resultant force
\therefore {\text{ Resultant force }}F{\text{ per unit length }} = {\text{ }}{F_1} + {F_2}
{\text{ }} = \dfrac{{24{\mu _o}}}{{2\pi r}} + \dfrac{{12{\mu _o}}}{{4\pi r}}
{\text{ }} = \dfrac{{60{\mu _o}}}{{4\pi r}} = \dfrac{{15{\mu _o}}}{{\pi r}}
\because {\text{ }}{{\text{F}}_1}{\text{ and }}{{\text{F}}_2}{\text{ both acting towards left therefore resultant force also acting }}{\text{towards left on the wire R}}
Hint: The resultant magnetic field is the vector sum of {B_x} and {B_y} .
Correct Option: C
Explanation for Correct Option:
Step 1: Find the magnetic field due to the half ring wire,
Magnetic field due to the half ring wire, {B_1} = \dfrac{{{\mu _o}i}}{{4R}} ,
where i is the current through the wire and R is the radius of the wire.
Step 2: Magnetic Field induction at point O due to the two straight wires,
Magnetic field at O, {B_2} = 2[\dfrac{{{\mu _o}i}}{{4\pi R}}(\cos 0 - \cos 90)]
Step 3: Find the resultant magnetic field at point O.
Resultant magnetic field,
B = \sqrt {{B_1}^2 + {B_2}^2}
B = \dfrac{{{\mu _o}i}}{{4\pi R}}{({\pi ^2} + {2^2})^{\dfrac{1}{2}}}
So, the magnetic induction at point P is B = \dfrac{{{\mu _o}i}}{{4\pi R}}{({\pi ^2} + {2^2})^{\dfrac{1}{2}}}
Two circular concentric loops of radii r_1=20cm and r_2 =30 cm are placed in the X-Y plane as shown in the figure. A current I=7 A is flowing through them. The magnetic moment of this loop system is
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