Correct Answer is Option (B)

P is the mid-point of line AC. A and C are the center of the two circle of each radius R.

Current flows through loop A and B are in same direction, So the magnetic field will also be in the same direction. Magnitude of magnetic field at paint P

= magnitude of magnetic field due to A and B at paint P.

= 2 $\left[\frac{{\mu }_{0}NI{R}^2}{2{\left(R^2+\frac{R^2}{4}\right)}^{\frac{3}{2}}}\right]$

$=$ $\frac{{\mu }_{0}NI{R}^2}{\frac{5^{\frac{3}{2}}{R}^3}{8}}$

= $\frac{8{\mu }_{0}NI}{5^{\frac{3}{2}}R}$

Correct Answer is Option (B)

Given situation is shown in the figure

As we know,

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

$\overrightarrow{F}=-q_0(\overrightarrow{v}\times \overrightarrow{B})$

According to question, direction of current is parallel to the force acting on the electron. Hence, the motion of test charge is towards the wire.

Correct Answer is Option (B)

Case (a) :



$B_A=\frac{{\mu }_0}{4\pi }\frac{I}{R}\times 2\pi$

$=\frac{{\mu }_0}{4\pi }\frac{I}{\ell /2\pi }\times 2\pi$ $$ $\left(2\pi R=\ell \right)$

$=\frac{{\mu }_0}{4\pi }\frac{I}{\ell }\times {\left(2\pi \right)}^2$

Case (b) :



$B_B=4\times \frac{{\mu }_0}{4\pi }\frac{I}{a/2}$ $\left[\sin {45}^{\circ }+\sin {45}^{\circ }\right]$

$=4\times \frac{{\mu }_0}{4\pi }\times \frac{I}{\ell /8}\times \frac{2}{\sqrt{2}}$

$=\frac{{\mu }_{0}I}{4\pi \ell }\times \sqrt[32]{2}\left[4a=1\right]$

Correct Answer is Option (A)

The magnetic field due a disc is given as

$B=\frac{h_0\omega Q}{2\pi R}$ i.e., $B\propto \frac{1}{R}$

Correct Answer is Option (D)

Current in a small element, $dl=\frac{d\theta }{\pi }I$

Magnetic field due to the element

$dB=\frac{{\mu }_0}{4\pi }\frac{2dl}{R}$

The component $dB$ $\cos \theta ,$ of the field is canceled by another opposite component.

Therefore,



$B_{net}=\int dB\sin \theta =\frac{{\mu }_{0}I}{2{\pi }^2{R}_0}$

$\underset{0}{\overset{\pi }{\int }}\sin \theta d\theta =\frac{{\mu }_{0}I}{{\pi }^{2}R}$