Question

An electron moving with a speed of 4.0 * 105 m>s in the positive x direction experiences zero magnetic force. When it moves in the positive y direction, it experiences a force of 3.2 * 10-13 N that points in the positive z direction. What are the direction and magnitude of the magnetic field

Answer 1

Answer:

B= 4.99T in negative x direction

Explanation:

F= qvBsin∅

q: charge on particle= 1.602 × 10^-19

v: velocity of the particle= 4 × 10⁵

B: Magentic field=???

∅: angle between magnetic field and direction of motion of charge particle

When moving in positive x- direction, the force is zero. According to the equation, the force will be zero when angle between direction of motion and magnetic field is either zero or 180⁰.

Now, when moving in positive y-direction, the force is 3.2 ˣ 10^-13 in positive z direction. and ∅=90. From Flemming's left hand rule, we can deduce that magnetic field is in negative x direction

3.2 ˣ 10^-13= 1.602 × 10^-19 × 4 × 10⁵ × B × sin90

B= 4.99T

Answer 2

Answer:

Explanation:

Given that,

Charge on an electron is

q=1.609×10^-19C

Speed of electron V=4×10^5 m/s in positive x direction

V = 4×10^5 •i m/s

At this point F=0N

When the electron is moving in positive y direction

V =4×10^5 •j m/s

F = 3.2×10^-13 N in positive z direction.

F = 3.2×10 ^-13 •k N

What is the direction and magnitude of the magnetic field B?

The force in a magnetic field is given as

F = q (V ×B)

The only time V×B is zero is when both V and B are in same direction

This shows that B is in the positive x direction or negative x direction

i.e B= Bx •i

Now applying this to the second condition

F= q(V×B)

3.2×10^-13 •k =1.609×10^-19 × (4×10^5•j ×Bx •i)

Note that, j×i =-k

3.2×10^-13 •k / 1.609×10^-19 = (4×10^5Bx •(j×i))

1988812.927 •k =- 4×10^5Bx•k

1988812.927/4×10^5 •k =-Bx •k

4.972 •k =-Bx •k

Then, Bx = -4.972 •k Telsa

Then, the magnitude of B is

B=4.972 T

And its direction is negative z direction.