Level 3

Biot-Savart law indicates that the moving electrons (velocity $v$) produce a magnetic field $B$ such that:

A proton is found to be accelerating on a cyclotron which has an oscillating frequency of $11 \mathrm{MHz}$ in an external magnetic field of $1 \mathrm{T}$. It is given that the radius of its dees is $55 \mathrm{cm}$. What is its kinetic energy (in $\mathrm{MeV}$)? ($m_{\mathrm{p}}=1.67\times {10}^{-27}\mathrm{ }\mathrm{k}\mathrm{g}\text{,}\mathrm{ }e=1.6\times \mathrm{ }{10}^{-19}\mathrm{ }\mathrm{C}$)

Three long straight wires $A\text{,} B$ and $C$ are carrying currents as shown in the figure. The resultant force on $B$ is directed,

Two $\mathrm{\alpha }$–particles enter a region of uniform magnetic field with their velocities in the ratio $3:2$. Calculate the ratio of the respective radii of their paths.

The magnitude of magnetic induction at a distance $4 \mathrm{cm}$ due to straight conductor carrying a current of $10 \mathrm{A}$ is,

In vector form, the magnetic induction $dB$ at a point of distance $r$ from centre of element of length $dl$ is given as,

If the current flowing in a circular loop is in clockwise direction, then the magnetic induction will be,

An electron moves in a circular orbit with a uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is proportional to,