Answer :
The answer is 2.65*10^18 m/s^2
The step-wise calculation is-
a=Fb/me
=e(vXB)/me
=−1.6*10^−19(6.0*10i×(3i+1.5j+2k))/9.1*10^−31
=−1.05*10^18(1.5k−2j)
Now, |a|=1.05*10^18√1.5^2+(−2)^2=2.65*10^18 m/s^2
How do you find the acceleration of a particle in a magnetic field?
- The acceleration of a particle in a circular orbit is: Using F = ma,
- Thus the radius of the orbit depends on the particle's momentum, mv, and
- the product of the charge and strength of the magnetic field.
Magnetic field magnitude = (permeabilityoffreespace)/(currentmagnitude)2π(distance)
Learn more about the acceleration of a particle in a magnetic field here
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The magnitude of the acceleration of the electron at this point |a| = 2.64*10^18 m/sec^2
Given electron velocity = v = 6.0 * 10 ^ 6 m / sec. In the positive x direction, therefore v = 6 * 10 ^ 6 i
, that is, magnetic field = B = Bx i + By j + Bz k
B = 3.0 i + 1.5 j + 2.0 k
Now, The force applied to the electrons in the magnetic field is given by
Fm = q * (vXB)
Here, q = electron charge = 1.602 * 10 ^ -19 C
From Newton's law,
Fnet = m * a
Fm = m * a
Given m = electron mass = 9.1 * 10 ^ -31 kg
a = Acceleration = ??
Therefore, a = Fm / m = (q / m) * (vXB) = [(1.602 * 10 ^ -19) / (9.1 * 10 ^ -31)] * [(6.0 * 10 ^ 6) X (3.0i + 1.5j + 2.0k)]
a = [(1.602 * 10 ^ -19) / (9)
1 * 10^-31)] * [18.0 * 10^6 (ixi) + 9.0 * 10^6 (ixj) + 12.0 * 10^6 (ixk)]
Da, ixi = 0
ixj = k
ixk = -j
So a = [(1.602 * 10^-19) / (9.1 * 10^-31)] * [9.0 * 10^6 k-12, 0 * 10^6 j]
a = - [(1.602 * 10^-19) / (9.1 * 10^-31)] * 12 * 10^6 j + [(1.602 * 10^-19) / (9.1 * 10^-31)] * 9 * 10^6 k
a = -2.11 * 10^18 j + 1.58 * 10^18 k
So the acceleration magnitude = |a| = sqrt[(2.11 * 10^18)^2 + ( 1.58*10^18)^2]
|a| = 2.64*10^18 m/sec^2
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