Question 1:
moderateFrom Ampere’s circuital law for a long straight wire of circular cross section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is:
(2022)
Inside a uniform cylindrical wire, the magnetic field is directly proportional to the radius ($B \propto r$), increasing linearly. Outside the wire, the magnetic field is inversely proportional to the distance ($B \propto 1/r$). Therefore, option D is the correct choice.
Question 2:
easyA long solenoid of radius $1\text{ mm}$ has $100$ turns per mm. If $1\text{ A}$ current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:
(2022)
The magnetic field inside a long solenoid is given by $B = \mu_0 n I$. Given $n = 100\text{ turns/mm} = 10^5\text{ turns/m}$ and $I = 1\text{ A}$, substituting the values yields $B = (4\pi \times 10^{-7}) \times 10^5 \times 1 = 12.56 \times 10^{-2}\text{ T}$.
Question 3:
moderateA long solenoid of $50text{ cm}$ length having $100$ turns carries a current of $2.5text{ A}$. The magnetic field at the centre of the solenoid is:Â ($\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}$)
(2020)
The number of turns per unit length is $n = \frac{N}{L} = \frac{100}{0.5\text{ m}} = 200\text{ turns/m}$. Using the formula $B = \mu_0 n I$, we substitute $\mu_0 = 4\pi \times 10^{-7}$, $n = 200$, and $I = 2.5\text{ A}$ to get $B = 6.28 \times 10^{-4}\text{ T}$.
Question 4:
easyA long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is:
(2003)
The magnetic field of a solenoid is $B = mu_0 n I$. When the current is doubled ($I' = 2I$) and turns per unit length are halved ($n' = n/2$), the new magnetic field is $B' = mu_0 (n/2)(2I) = mu_0 n I = B$.