Correct Answer is Option (B)

We know, magnetic field at centre of solenoid

$B={\mu }_0\frac{N}{l}l={\mu }_{0}nl$ $\left[n=\frac{N}{l}\right]$

$=4\pi \times {10}^{-7}\times 100\times {10}^3\times 1$ $\left[n=\frac{100}{{10}^{-3}}\right]$

$=4\pi \times {10}^{-2}T$

$B=12.56\times {10}^{-2}T$

Correct Answer is Option (C)

For solid wire

Inside point

$B=\frac{{\mu }_{0}I{r}^2}{R^2\times 2\pi r}$

$=\frac{{\mu }_{0}Ir}{R^2\times 2\pi }$

$B\propto r$

Outside point

$B=\frac{{\mu }_{0}I}{2\pi r}$

$B\propto \frac{1}{r}$

Correct Answer is Option (D)

Magnetic field at centre of solenoid = ${\mu }_{0}nI$
$n=\frac{N}{L}=\frac{100}{50\times {10}^{-2}}$
= 200 turns/m
I = 25A
Now, substituting the values,
B = 4$\pi$ $\times$ 10^-7 $\times$ 200 $\times$ 2.5
= 6.28 $\times$ 10^-4 T

Correct Answer is Option (C)

The magnetic field inside conductor is given as

B = $\frac{{\mu }_0}{2\pi }\frac{i}{R^2}d$

$\therefore$ B $\propto$ d

If, straight line passing through origin, at surface (d = R)

B = $\frac{{\mu }_0}{2\pi }\frac{i}{R}$

For the maximum B at outside surface (d > R)

B = $\frac{{\mu }_0}{2\pi }\frac{i}{d}$

$\Rightarrow$ B $\propto$ $\frac{1}{d}$

Correct Answer is Option (A)

For points inside the wire i.e., (r $\le$ R)

Magnetic field $B=\frac{{\mu }_{0}Ir}{2\pi R^2}$

For points outside the wire (r $\ge$ R)

Magnetic field, $B^{\prime }=\frac{{\mu }_{0}I}{2\pi R}$

$\therefore \frac{B}{B^{\prime }}=\frac{\frac{{\mu }_{0}I\left(a/2\right)}{2\pi a^2}}{\frac{{\mu }_{0}I}{2\pi \left(2a\right)}}=1:1$

Correct Answer is Option (C)

Magnetic moment of the loop.

M = NIA = 2000 × 2 × 1.5 × 10^–4 = 0.6 J/T

Torque $\tau$ = MBsin30°

$=0.6\times 5\times {10}^{-2}\times \frac{1}{2}=1.5\times {10}^{-2}Nm$

Correct Answer is Option (B)

Magnetic field induction at point inside the solenoid of length l, having n turns per unit length carrying current i is given by

B = $\mu$₀ni

If i $\to$ doubled, n $\to$ halved then B $\to$ remains same.