A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will
A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will
- [AIPMT 1997]
- A
Get deflected vertically upwards
- B
Move in a circular orbit with its speed increased
- C
Move in a circular orbit with its speed unchanged
- D
Continue to move due east
Similar Questions
A magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of $0.75 \;T$. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through $15\; kV$ enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is $9.0 \times 10^{-5} \;V\, m ^{-1},$ make a simple guess as to what the beam contains. Why is the answer not unique?
A magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of $0.75 \;T$. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through $15\; kV$ enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is $9.0 \times 10^{-5} \;V\, m ^{-1},$ make a simple guess as to what the beam contains. Why is the answer not unique?
A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction, extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is
A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction, extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is
- [IIT 2002]
The radius of curvature of the path of the charged particle in a uniform magnetic field is directly proportional to
The radius of curvature of the path of the charged particle in a uniform magnetic field is directly proportional to
A proton (mass $m$ ) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B.$ The field occupies a region of space by width $'d'$. If $\alpha $ be the angle of deviation of proton from initial direction of motion (see figure), the value of $sin\,\alpha $ will be
A proton (mass $m$ ) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B.$ The field occupies a region of space by width $'d'$. If $\alpha $ be the angle of deviation of proton from initial direction of motion (see figure), the value of $sin\,\alpha $ will be
- [JEE MAIN 2015]
A electron experiences a force $\left( {4.0\,\hat i + 3.0\,\hat j} \right)\times 10^{-13} N$ in a uniform magnetic field when its velocity is $2.5\,\hat k \times \,{10^7} ms^{-1}$. When the velocity is redirected and becomes $\left( {1.5\,\hat i - 2.0\,\hat j} \right) \times {10^7}$, the magnetic force of the electron is zero. The magnetic field $\vec B$ is :
A electron experiences a force $\left( {4.0\,\hat i + 3.0\,\hat j} \right)\times 10^{-13} N$ in a uniform magnetic field when its velocity is $2.5\,\hat k \times \,{10^7} ms^{-1}$. When the velocity is redirected and becomes $\left( {1.5\,\hat i - 2.0\,\hat j} \right) \times {10^7}$, the magnetic force of the electron is zero. The magnetic field $\vec B$ is :