Proton with kinetic energy of $1\;MeV$ moves from south to north. It gets an acceleration of $10^{12}\; \mathrm{m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field :.......$mT$ (Rest mass of proton is $1.6 \times 10^{-27} \;\mathrm{kg}$ )

The magnetic moments associated with two closely wound circular coils $A$ and $B$ of radius $r_A=10 cm$ and $r_B=20 cm$ respectively are equal if: (Where $N _A, I _{ A }$ and $N _B, I _{ B }$ are number of turn and current of $A$ and $B$ respectively)

A charged particle moves through a magnetic field perpendicular to its direction. Then

A charged particle carrying charge $1\,\mu C$  is moving with velocity $(2 \hat{ i }+3 \hat{ j }+4 \hat{ k })\, ms ^{-1} .$ If an external magnetic field of $(5 \hat{ i }+3 \hat{ j }-6 \hat{ k }) \times 10^{-3}\, T$ exists in the region where the particle is moving then the force on the particle is $\overline{ F } \times 10^{-9} N$. The vector $\overrightarrow{ F }$ is :

A particle moving in a magnetic field increases its velocity, then its radius of the circle

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