An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic field with x component 0.041 T and y component -0.16 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.

This question involves the concepts of the magnetic field, triple-scalar product, and magnetic force.

(a) The magnitude of the magnetic force on the electron is " 0.876 x 10⁻¹³ N".

(b) The magnitude of the magnetic force on the proton is "-0.876 x 10⁻¹³ N".

(a)

The magnitude of a magnetic force applied on a charged particle due to the magnetic field is given by the following formula:

where,

F = magnetic force = ?

q = charge on electron = -1.6 x 10⁻¹⁹ (i + j + k) C

v = velocity = (2.6 i + 3.2 j +0 k) x 10⁶ m/s

B = magnetic field = (0.041 i - 0.16 j + 0 k) T

Therefore,

This can be solved using the scalar-triple product:

Expanding by column 3, we get:

F = 0.876 x 10⁻¹³ N

(b)

For the proton the charge changes to:

q = 1.6 x 10⁻¹⁹ (i + j + k) C

Therefore,

This can be solved using the scalar-triple product:

Expanding by column 3, we get:

F = - 0.876 x 10⁻¹³ N

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Attached picture shows the formula of magnetic force.

a) , b)

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

The charge of the electron is equal to . Then, cross product can be solved by using determinants:

The magnetic force is:

b) The charge of the proton is equal to . Then, cross product has the following determinant:

The magnetic force is:

the answer is b

explanation: