Question

maximum magnetic field on electron moving with a velocity 4×10 4 m/s when it moves in magnetic fields intensity 5×10 ​

Answer 1

Explanation:

q = 1.6×10^-10 C

v = 4×10^4 m/s

B = 5×10 T

F = ?

F = qvB

F = (1.6×10^-19) ×(4×10^4)×(5×10)

F= 3.2×10^-13 N

Answer 2

The maximum magnetic field on the electron will be 3.2 × $10^{-19}$ N.

Given:

The charge of the electron, e= 1.6 × $10^{-19}$ C

The velocity of the electron, v= 4 × $10^{4}$ m/s

The magnetic field intensity, B= 5 × $10^{-5}$ T

To Find:

The maximum magnetic field.

Solution:

From equation F =q(v ×B ), the maximum magnetic force on the electron

$F_{max} = 1.6 * 10^{-19}* 4 * 10^{4} * 5 * 10^{-5}$= 3.2×$10^{-19}$ N

Hence, the maximum magnetic field on the electron will be 3.2 × $10^{-19}$ N.

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