
A particle of mass \[0.5gm\] and charge \[2.5 \times {10^{ - 8}}C\] is moving with velocity \[6 \times {10^4}m/s\]. What should be the minimum value of the magnetic field acting on it, so that the particle is able to move in a straight line?
( \[g = 9.8m/{s^2}\] )
A) \[0.327Weber/{m^2}\]
B) \[3.27Weber/{m^2}\]
C) \[32.7Weber/{m^2}\]
D) None of these
Answer
570.6k+ views
Hint: Find the force acting on a moving charge when you are given charge, velocity and magnetic field. For the charge to be moved in a straight line a counter force must act, in this case this counter force is gravitational force. Now equate the gravitational force with the force on a moving charge under constant magnetic field.
Complete step by step answer:
Magnetic fields exert forces on moving charges. The direction of the magnetic force on a moving charge is perpendicular to the plane formed by \[v\] and \[B\] and follows right hand rule. The magnitude of the force is proportional to \[q\] , \[v\] , \[B\] and the \[\sin \theta \] of the angle between \[v\] and \[B\] .
Now in the given question \[B\] and \[v\] are perpendicular therefore, \[\sin \theta \] will be 1.
Now equation force due to acceleration with force on a moving charge under constant magnetic field.
Force due to gravity \[ = mg = 0.5 \times {10^{ - 3}} \times 9.8\]
\[F = 4.5 \times {10^{ - 3}}N\] --(1)
Force due charge = \[ = q|v \times B| = 2.5 \times {10^{ - 8}}(6 \times {10^4} \times B)\] --(2)
Now equating equation 1 and 2.
\[4.5 \times {10^{ - 3}} = 2.5 \times {10^{ - 8}}(6 \times {10^4} \times B)\]
\[B = \dfrac{{4.5 \times {{10}^{ - 3}}}}{{2.5 \times {{10}^{ - 8}}(6 \times {{10}^4})}}\]
\[B = 0.327Weber/{m^2}\]
Hence, Option A is correct.
Note: No magnetic force acts on static charges. Although, there is a magnetic force on moving charges. When charges are at rest, their electric fields do not affect magnets. But, when charges move, they produce magnetic fields which also exert forces on other magnets in vicinity. When there is relative motion, a connection between electric and magnetic fields emerges in which each affects the other.
Complete step by step answer:
Magnetic fields exert forces on moving charges. The direction of the magnetic force on a moving charge is perpendicular to the plane formed by \[v\] and \[B\] and follows right hand rule. The magnitude of the force is proportional to \[q\] , \[v\] , \[B\] and the \[\sin \theta \] of the angle between \[v\] and \[B\] .
Now in the given question \[B\] and \[v\] are perpendicular therefore, \[\sin \theta \] will be 1.
Now equation force due to acceleration with force on a moving charge under constant magnetic field.
Force due to gravity \[ = mg = 0.5 \times {10^{ - 3}} \times 9.8\]
\[F = 4.5 \times {10^{ - 3}}N\] --(1)
Force due charge = \[ = q|v \times B| = 2.5 \times {10^{ - 8}}(6 \times {10^4} \times B)\] --(2)
Now equating equation 1 and 2.
\[4.5 \times {10^{ - 3}} = 2.5 \times {10^{ - 8}}(6 \times {10^4} \times B)\]
\[B = \dfrac{{4.5 \times {{10}^{ - 3}}}}{{2.5 \times {{10}^{ - 8}}(6 \times {{10}^4})}}\]
\[B = 0.327Weber/{m^2}\]
Hence, Option A is correct.
Note: No magnetic force acts on static charges. Although, there is a magnetic force on moving charges. When charges are at rest, their electric fields do not affect magnets. But, when charges move, they produce magnetic fields which also exert forces on other magnets in vicinity. When there is relative motion, a connection between electric and magnetic fields emerges in which each affects the other.
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