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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