Questions in Moving Charges and Magnetism

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The field due to a long straight wire carrying a current Iis proportional to
Two concentric coils each of radius equal to $2\pi {\rm{ }}cm$ are placed at right angles to each other. 3 ampere and 4 ampere are the currents flowing in each coil respectively. The magnetic induction in $Weber/{m^2}$ at the centre of the coils will be $({\mu _0} = 4\pi \times {10^{ - 7}}Wb/A.m)$
A wire carrying current I and other carrying 2I in the same direction produces a magnetic field B at the mid point. What will be the field when 2I wire is switched off
Two long parallel wires P and Q are both perpendicular to the plane of the paper with distance 5 m between them. If P and Q carry current of 2.5 amp and 5 amp respectively in the same direction, then the magnetic field at a point half way between the wires is
The direction of magnetic lines of force produced by passing a direct current in a conductor is given by
For the magnetic field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be
A proton moving with a constant velocity passes through a region of space without any change in its velocity. If $\vec E $ and $\vec B $ represent the electric and magnetic fields respectively, then this region of space may have
A uniform electric field and a uniform magnetic field are produced, pointed in the same direction. An electron is projected with its velocity pointing in the same direction
Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describes circular path of radius ${R_1}$ and ${R_2}$ respectively. The ratio of mass of X to that of Y is
A beam of ions with velocity $2 \times {10^5}\,m/s$ enters normally into a uniform magnetic field of $4 \times {10^{ - 2}}\,tesla$. If the specific charge of the ion is $5 \times {10^7}\,C/kg$, then the radius of the circular path described will be

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