Two mutually perpendicular insulated conducting wires carrying equal currents $I$ intersect at the origin. The resultant magnetic induction at point $P(2 \ m, 3 \ m)$ will be:

$A$ long solenoid has $100 \, \text{turns/m}$ and carries current $i$. An electron moves within the solenoid in a circle of radius $2.30 \, \text{cm}$ perpendicular to the solenoid axis. The speed of the electron is $0.046 \, c$ ($c = 3 \times 10^8 \, \text{m/s}$ is the speed of light). Find the current $i$ in the solenoid (approximate). (in $ \text{A}$)

Find the magnetic field at the point $P$ in the figure. The curved portion is a quarter-circle connected to two long straight wires.

$A$ particle having a charge $50 \ e$ is revolving in a circular path of radius $0.4 \ m$ with $1 \ r.p.s.$ The magnetic field produced at the centre of the circle is $(\mu_0 = 4 \pi \times 10^{-7} \ SI \ units$ and $e = 1.6 \times 10^{-19} \ C)$.

$A$ long straight wire carries a current of $35\, A$. What is the magnitude of the magnetic field $B$ at a point $20\, cm$ from the wire?

$A$ particle carrying a charge equal to $1000$ times the charge on an electron is rotating at $1$ rotation per second in a circular path of radius $r \ m$. If the magnetic field produced at the centre of the path is $x$ times the permeability of vacuum $\mu_0$,the radius $r$ in $m$ is: $[e = 1.6 \times 10^{-19} \ C], [x = 2 \times 10^{-16}]$