A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 6.75 × 10 − 16

Question

A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 6.75 × 10 − 16 N 6.75×10−16 N as it moves at 207 m/s 207 m/s at 74.9 ∘ 74.9∘ to the direction of the field. Find the magnitude of the magnetic field.

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Thiên Hương 3 years 2021-07-27T12:49:03+00:00 1 Answers 20 views 0

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  1. Edana
    0
    2021-07-27T12:50:50+00:00
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    Answer:

    The magnitude of the magnetic field is 10.19 T.

    Explanation:

    Given that,

    Charge on the doubly charged molecule, q = 2e

    Magnetic force, F=6.75\times 10^{-16}\ N

    Speed of molecule, v = 207 m/s

    The angle to the direction of the field is 74.9 degrees.

    We need to find the magnitude of magnetic field. The formula of magnetic force is given by :

    F=qvB

    B is magnetic field

    B=\dfrac{F}{qv}\\\\B=\dfrac{6.75\times 10^{-16}}{2\times 1.6\times 10^{-19}\times 207}\\\\B=10.19\ T

    So, the magnitude of the magnetic field is 10.19 T.

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