An electron (mass 9 × 10⁻³¹ kg and charge 1.6 × 10⁻¹⁹ C) moving with speed c/100 (c = speed of light) is injected into a magnetic field of magnitude 9 × 10⁻⁴ T perpendicular to its direction of motion. We wish to apply an uniform electric together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3 × 10⁸ ms⁻¹)

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An electron (mass 9 × 10⁻³¹ kg and charge 1.6 × 10⁻¹⁹ C) moving with speed c/100 (c = speed of light) is injected into a magnetic field of magnitude 9 × 10⁻⁴ T perpendicular to its direction of motion. We wish to apply an uniform electric together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3 × 10⁸ ms⁻¹)

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is perpendicular to and
its magnitude is 27 × 10⁴ V m⁻¹
is perpendicular to and
its magnitude is 27 × 10² V m⁻¹
is parallel to and
its magnitude is 27 × 10² V m⁻¹
is parallel to and
its magnitude is 27 × 10⁴ V m⁻¹

Answer 1

Calculation:
The magnetic force on a moving charge is given by:
F_magnetic = q(v × B)
The electric force is given by:
F_electric = qE
For no deflection, the forces must balance:
F_electric = F_magnetic
qE = qv × B
Therefore, E = v× B which means E should be perpendicular to both v and B.
Now, the velocity (v) of the electron is:
v = c / 100 = (3 × 10⁸) / 100 = 3 × 10⁶ m/s
Substitute the values for B and v:
E = (3 × 10⁶) × (9 × 10^-4) = 27 × 10² V/m
Conclusion: The electric field (E) is perpendicular to the magnetic field (B) and its magnitude is 27 × 10² V/m.

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