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Question
A particle of mass m and charge q is thrown from origin at t=0 with velocity 2^i+3^j+4^k m/s in a region with uniform magnetic field →B=2^i T. After time t=πmqB sec, an electric field is switched on such that the particle moves on a straight line path with a constant speed. Then the electric field can be expressed as :
(B is the magnitude of magnetic field )
A particle of mass m and charge q is thrown from origin at t=0 with velocity 2^i+3^j+4^k m/s in a region with uniform magnetic field →B=2^i T. After time t=πmqB sec, an electric field is switched on such that the particle moves on a straight line path with a constant speed. Then the electric field can be expressed as :
(B is the magnitude of magnetic field )
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Solution
The correct option is D 8^j−6^k V/m
The particle is moving with initial velocity,
→vi=2^i+3^j+4^k m/s
By using the relation for magnetic force (−→Fm=q(→v×→B)), we can find the force acting in YZ plane.
The magnetic field is →B=2^i T
Here we can see that the time interval is,
t=πmqB=T2
Hence, after t=T2 i.e. on completing the half revolution the components of velocity perpendicular to the direction of magnetic field will get reversed, without change in the speed.
Thus, at t=πmqB, the velocity of the particle will be;
→v=2^i−3^j−4^k
Now the particle continue to move in a straight line with a constant speed as soon as the electric field is switched on.
→v=constant, or →a=0
Thus, −−→Fnet=0
⇒q→E+q(→v×→B)=0
→E=−(→v×→B)
→E=−[(2^i−3^j−4^k)×(2^i)]
→E=−[0+(6^k)+(−8^j)]
∴→E=−6^k+8^j=8^j−6^k V/m
Hence, option (d) is the correct answer.
The particle is moving with initial velocity,
→vi=2^i+3^j+4^k m/s
By using the relation for magnetic force (−→Fm=q(→v×→B)), we can find the force acting in YZ plane.
The magnetic field is →B=2^i T
Here we can see that the time interval is,
t=πmqB=T2
Hence, after t=T2 i.e. on completing the half revolution the components of velocity perpendicular to the direction of magnetic field will get reversed, without change in the speed.
Thus, at t=πmqB, the velocity of the particle will be;
→v=2^i−3^j−4^k
Now the particle continue to move in a straight line with a constant speed as soon as the electric field is switched on.
→v=constant, or →a=0
Thus, −−→Fnet=0
⇒q→E+q(→v×→B)=0
→E=−(→v×→B)
→E=−[(2^i−3^j−4^k)×(2^i)]
→E=−[0+(6^k)+(−8^j)]
∴→E=−6^k+8^j=8^j−6^k V/m
Hence, option (d) is the correct answer.
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