
A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will
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
Answer
163.5k+ views
Hint: If a charged particle is moving in a magnetic field then due to motion of the charged particle and magnetic field interaction there is deflection in the path of the motion of the charged particle. The magnitude of the deflection of the particle is proportional to the magnitude of the magnetic force acting.
Formula used:
\[\overrightarrow F = q\overrightarrow E + q\left( {\overrightarrow v \times \overrightarrow B } \right)\], here F is the net force acting on the particle of charge q moving with speed v in a region with magnetic field B and electric field E.
Complete answer:
In the region, there is only magnetic field and the electric field is absent. So, the force acting on the positively charged particle is,
\[\vec F = + q\left( {\vec v \times \vec B} \right)\]
It is given that the particle is moving due to east.
Considering the direction towards east as equivalent to the +x-axis of Cartesian coordinate system, the direction vertically upward will be +z-axis.
Let the speed is \[{v_0}\]and the magnetic field strength is \[{B_0}\]
Then he velocity vector is \[\vec v = {v_0}\hat i\]and the magnetic field vector is \[\vec B = {B_0}\hat k\]
Putting in the equation for the magnetic force, we get
\[\vec F = + q\left( {{v_0}\hat i \times {B_0}\hat k} \right)\]
\[\vec F = + q{v_0}{B_0}\left( { - \hat j} \right)\]
\[\vec F = - \left( {q{v_0}{B_0}} \right)\hat j\]
So, the magnetic force is acting in the direction of negative y-axis, i.e. along the geographical south.
As the magnetic force is acting perpendicular to the initial direction of motion, so the speed of the particle will be unchanged and the path of the motion will be circular.
Therefore, the correct option is (C).
Note:We should be careful while making the directional equivalence of the geographical directions to the Cartesian coordinate system. We should maintain one specific coordinate system throughout the solution.
Formula used:
\[\overrightarrow F = q\overrightarrow E + q\left( {\overrightarrow v \times \overrightarrow B } \right)\], here F is the net force acting on the particle of charge q moving with speed v in a region with magnetic field B and electric field E.
Complete answer:
In the region, there is only magnetic field and the electric field is absent. So, the force acting on the positively charged particle is,
\[\vec F = + q\left( {\vec v \times \vec B} \right)\]
It is given that the particle is moving due to east.
Considering the direction towards east as equivalent to the +x-axis of Cartesian coordinate system, the direction vertically upward will be +z-axis.
Let the speed is \[{v_0}\]and the magnetic field strength is \[{B_0}\]
Then he velocity vector is \[\vec v = {v_0}\hat i\]and the magnetic field vector is \[\vec B = {B_0}\hat k\]
Putting in the equation for the magnetic force, we get
\[\vec F = + q\left( {{v_0}\hat i \times {B_0}\hat k} \right)\]
\[\vec F = + q{v_0}{B_0}\left( { - \hat j} \right)\]
\[\vec F = - \left( {q{v_0}{B_0}} \right)\hat j\]
So, the magnetic force is acting in the direction of negative y-axis, i.e. along the geographical south.
As the magnetic force is acting perpendicular to the initial direction of motion, so the speed of the particle will be unchanged and the path of the motion will be circular.
Therefore, the correct option is (C).
Note:We should be careful while making the directional equivalence of the geographical directions to the Cartesian coordinate system. We should maintain one specific coordinate system throughout the solution.
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