Moving Charges and Magnetism Set-5

Q1. A proton of charge $+e$ and mass $m_p=1.67\times10^{-27}\,\mathrm{kg}$ moves with speed $v=2.0\times10^{6}\,\mathrm{m/s}$ perpendicular to a uniform magnetic field $B=1.0\times10^{-2}\,\mathrm{T}$. The radius of its circular path is

Q2. A long straight wire along the $z$‑axis carries steady current $I$ in the $+z$ direction. At point $P$ on the $+x$ axis a particle of charge $q$ moves with velocity $\vec v=v\,\hat{z}$. The magnetic force $\vec F$ on the particle due to the wire's field is

Q3. A proton ($m_p=1.67\times10^{-27}\,\mathrm{kg}$, charge $+e$) with speed $v=3.0\times10^{6}\,\mathrm{m/s}$ enters a region of uniform magnetic field $B=1.0\times10^{-2}\,\mathrm{T}$ directed into the page. The field region extends a width $d=0.50\ \mathrm{m}$ along the proton's initial direction. After leaving the field the proton travels a further distance $L=1.00\ \mathrm{m}$ to strike a screen. Neglecting gravity, the perpendicular deflection on the screen is approximately

Q4. A rectangular loop of wire of height $l$ (vertical) and width $w$ (horizontal) lies in the plane of the page. The region $x>0$ has a uniform magnetic field $\vec B=-B\,\hat{z}$ (into the page). The loop (resistance $R$) is pulled to the right with constant speed $v$ so its leading edge is entering the $B$‑region. At the instant when part of the loop lies inside the field, the magnitude of the external force required to keep the speed constant is

Q5. In a Thomson‑type velocity‑selector followed by magnetic deflection, ions first pass through crossed fields $E_1=2.0\times10^{4}\,\mathrm{V/m}$ and $B_1=0.10\ \mathrm{T}$, so selected ions have $v=E_1/B_1$. They then enter a region with perpendicular magnetic field $B_2=0.200\ \mathrm{T}$ and move in circular paths of radius $r=0.150\ \mathrm{m}$. The mass‑to‑charge ratio $m/q$ of the selected ions equals

Q6. A proton (mass $1.67\times10^{-27}\,\mathrm{kg}$, charge $1.6\times10^{-19}\,\mathrm{C}$) has kinetic energy $3.2\times10^{-13}\,\mathrm{J}$ and enters a uniform magnetic field $B=0.5\,\mathrm{T}$ with velocity perpendicular to the field. Using $v=\sqrt{\dfrac{2K}{m}}$ and $r=\dfrac{mv}{qB}$, the radius of its circular path is:

Q7. An electron (mass $9.11\times10^{-31}\,\mathrm{kg}$, charge $1.6\times10^{-19}\,\mathrm{C}$) with velocity components $v_x=2.0\times10^{6}\,\mathrm{m/s}$ and $v_z=3.0\times10^{6}\,\mathrm{m/s}$ enters a region of uniform magnetic field $B=0.01\,\mathrm{T}$ directed along $+z$. The motion is helical. The pitch $p$ (distance advanced along $z$ in one complete turn) is given by $p=v_{\parallel}T$ with $T=\dfrac{2\pi m}{|q|B}$. The pitch equals:

Q8. A long straight cylindrical conductor of radius $R$ carries total current $I$ distributed so that the current density varies as $J(r)=k\,r$ (where $r$ is radial distance from the axis). For $r<R$, the magnetic field $B(r)$ inside the conductor is:

Q9. Three infinitely long straight parallel conductors are perpendicular to the plane and placed at the vertices of an equilateral triangle of side $a$. Two carry current $I$ out of the plane and the third carries current $I$ into the plane. The magnitude of the resultant magnetic field at the centroid is:

Q10. A conducting slab of thickness $t=1.0\times10^{-3}\,\mathrm{m}$ carries a current $I=10\,\mathrm{A}$. A magnetic field $B=1.0\,\mathrm{T}$ is applied perpendicular to the slab and a Hall voltage $V_H=0.5\times10^{-3}\,\mathrm{V}$ is measured across the thickness. Assuming charge carriers are electrons with magnitude $e=1.6\times10^{-19}\,\mathrm{C}$, the carrier number density $n$ (using $V_H=\dfrac{IB}{n e t}$) is:

Q11. A proton of mass $m_p=1.67\times10^{-27}\,\mathrm{kg}$ and charge $e=1.6\times10^{-19}\,\mathrm{C}$ is injected with speed $v=2.00\times10^{6}\,\mathrm{m/s}$ perpendicular to a uniform magnetic field $B=0.50\,\mathrm{T}$. Using $r=\dfrac{mv}{qB}$, the radius of its circular path is:

Q12. A proton moves in a region of uniform magnetic field $B$ perpendicular to its velocity. Initially its kinetic energy is $K$ and it moves in a circle of radius $r$. After passing through a thin foil it loses half of its kinetic energy (magnetic field remains unchanged). The new radius of curvature is:

Q13. A metallic slab of thickness $t=1.0\times10^{-3}\,\mathrm{m}$ carries a current $I=5.0\ \mathrm{A}$. A magnetic field $B=0.50\ \mathrm{T}$ perpendicular to the slab produces a Hall voltage $V_H=2.0\times10^{-4}\ \mathrm{V}$ across the thickness. If charge carriers have charge magnitude $q=1.6\times10^{-19}\ \mathrm{C}$, the carrier concentration $n$ (use $V_H=\dfrac{IB}{nqt}$) is approximately:

Q14. A circular loop of radius $R$ lies in the $xy$-plane with centre at the origin and carries current $I$ flowing anticlockwise when viewed from the positive $z$-axis. The magnetic field along the $z$-axis is non-uniform: $B(z)=B_0+\alpha z$ with constant $\alpha$. The net force on the loop is:

Q15. A charged particle (mass $m$, charge $q$) moves in a uniform magnetic field $B$ with velocity components $v_{\perp}$ (perpendicular to $B$) and $v_{\parallel}$ (parallel to $B$). The helical radius and pitch are $r=\dfrac{mv_{\perp}}{qB}$ and $p=v_{\parallel}\dfrac{2\pi m}{qB}$. If the magnetic field is changed to $2B$ while $v_{\perp}$ is simultaneously changed to $2v_{\perp}$ (but $v_{\parallel}$ remains unchanged), the new radius $r'$ and pitch $p'$ are: