Correct Answer is Option (A)

Magnetic fields due to the two parts at their common centre are respectively,

$B_y=\frac{{\mu }_{0}i}{4R}$ and $B_z=\frac{{\mu }_{0}i}{4R}$


Resultant field = $\sqrt{B_y^2+B_z^2}$

= $\sqrt{{\left(\frac{{\mu }_{0}i}{4R}\right)}^2+{\left(\frac{{\mu }_{0}i}{4R}\right)}^2}$

$=\sqrt{2}\frac{{\mu }_{0}i}{4R}=\frac{{\mu }_{0}i}{2\sqrt{2}R}$

Correct Answer is Option (B)

When the ring rotates about its axis with a uniform frequency f Hz, the current flowing in the ring is

I = $\frac{q}{T}$ = qf

Magnetic field at the centre of the ring is

B = $\frac{{\mu }_{0}I}{2R}$ = $\frac{{\mu }_{0}qf}{2R}$

Correct Answer is Option (C)

Let r₁ and r₂ are the radius of coil 1 and 2. If B₁ and B₂ are magnetic induction at their centre, then

$B_1=\frac{{\mu }_0{I}_1}{2r_1}$ and $B_2=\frac{{\mu }_0{I}_2}{2r_2}$

Since B₁ = B₂ ; and r₁ = 2r₂ therefore I₁ = 2I₂.

Again if R₁ and R₂ are resistance of the coil 1 and 2 then R₁ = 2R₂ (as R $\propto$ length = 2$\pi$r) and if V₁ and V₂ are the potential difference across them respectively, then

$\frac{V_1}{V_2}=\frac{I_1{R}_1}{I_2{R}_2}=\frac{\left(2I_2\right)\left(2R_2\right)}{I_2{R}_2}=4$

Correct Answer is Option (C)

B₁ = B = $\frac{{\mu }_{0}I}{2R}$

Further, B₂ = $\frac{{\mu }_0\left(2I\right)}{2r}$

Now 2 × 2$\pi$r = 2$\pi$R or r = R/2

Hence B₂ = $4\times \frac{{\mu }_{0}I}{2R}=4B$

Correct Answer is Option (C)

Magnetic field due to 5A = $\frac{5{\mu }_0}{2\pi \times 2.5}$ = $\frac{2{\mu }_0}{2\pi }\otimes$

Magnetic field due to 2.5A = $\frac{2.5{\mu }_0}{2\pi \times 2.5}$ = $\frac{{\mu }_0}{2\pi }\odot$

Resultant Magnetic field = $\frac{2{\mu }_0}{2\pi }-\frac{{\mu }_0}{2\pi }=\frac{{\mu }_0}{2\pi }\otimes$