$A$ coil wrapped around a toroid has an inner radius of $20 \,cm$ and an outer radius of $25 \,cm$. If the wire wrapping makes $800$ turns and carries a current of $12 \,A$, the maximum and minimum values of the magnetic field within the toroid are:

An imaginary north pole of $10 \, Am$ is rotating around an infinitely long current-carrying wire at $30 \, \text{revolutions/min}$ on a circular path. If the current in the wire is $5 \, A$, then calculate the work done in one second.

$A$ long solenoid is formed by winding $20$ $turns/cm$. The current necessary to produce a magnetic field of $20$ $mT$ inside the solenoid will be approximately ..... $A$ $(\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A})$.

$A$ 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

$A$ tightly wound long solenoid carries a current of $1.5 \text{ A}$. An electron is executing uniform circular motion inside the solenoid with a time period of $75 \text{ ns}$. The number of turns per metre in the solenoid is . . . . . . .
[Take mass of electron $m_e = 9 \times 10^{-31} \text{ kg}$,charge of electron $|q_e| = 1.6 \times 10^{-19} \text{ C}$,$\mu_0 = 4\pi \times 10^{-7} \text{ N/A}^2$,$1 \text{ ns} = 10^{-9} \text{ s}$]

$A$ solenoid of length $0.5\; m$ has a radius of $1\; cm$ and is made up of $500$ turns. It carries a current of $5\; A$. What is the magnitude of the magnetic field inside the solenoid?