Given that the speed of the particle is 4 \hat{a}_x m/s in a region where E = 20 \hat{a}_y V/m and B = B_0 \hat{a}_z wb/m^2, determine the value of B_0 such that the velocity of the particle remains constant.
Answer (2)
To keep the particle's speed constant at 4 m/s, the magnetic field strength B 0 must be − 5 Wb/m\u00b2. This ensures that the forces from the electric and magnetic fields balance each other out. In vector terms, the resultant force acting on the particle is zero, leading to uniform motion.
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To solve this problem, we need to find the magnetic field strength B 0 such that the velocity of the particle remains constant in a region with an electric field E = 20 a ^ y V/m and a magnetic field B = B 0 a ^ z Wb/m².
The Lorentz force acting on a charged particle with charge q , moving with velocity v , in the presence of an electric field E and magnetic field B , is given by:
F = q ( E + v × B )
Since we want the velocity of the particle to remain constant, the net force should be zero ( F = 0 ). This implies that:
E + v × B = 0
Let's identify our variables:
The velocity v = 4 a ^ x m/s.
The electric field E = 20 a ^ y V/m.
The magnetic field B = B 0 a ^ z Wb/m².
We need to compute the cross product v × B :
v × B = ( 4 a ^ x ) × ( B 0 a ^ z ) = 4 B 0 ( a ^ x × a ^ z )
Using the right-hand rule, a ^ x × a ^ z = − a ^ y , so:
v × B = − 4 B 0 a ^ y
Setting the net force to zero:
E + v × B = 20 a ^ y − 4 B 0 a ^ y = 0
Solving for B 0 :
20 − 4 B 0 = 0 4 B 0 = 20 B 0 = 5
Thus, the value of B 0 must be 5 Wb/m² to ensure that the velocity of the particle remains constant.
