A proton moves with a velocity
V (2i - 4 j + k) m / s
in a region where the field is given by
B = (i + 2 j -3k) T
. What is the magnitude of the magnetic force experienced by the load?
To find the magnitude of the magnetic force experienced by the proton, we can use the formula for the magnetic force on a moving charged particle in a magnetic field. The formula is:
F = q * (v x B)
where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field vector.
In this case, the proton has a charge of +e (elementary charge) and a velocity of V = (2i - 4j + k) m/s. The magnetic field vector is given as B = (i + 2j - 3k) T.
Now, let's calculate the cross product of v and B:
v x B = (2i - 4j + k) x (i + 2j - 3k)
The cross product of two vectors can be calculated by determining the determinant of a 3x3 matrix. The resulting cross product vector is obtained by subtracting the products of the diagonals in one direction from the products of the diagonals in the other direction. The cross product is determined for each component: x, y, and z.
So, let's calculate the cross product:
i x i = 0; i x 2j = 2k; i x -3k = -3j
4j x i = -4k; 4j x 2j = 0; 4j x -3k = -12i
-k x i = -j; -k x 2j = 2i; -k x -3k = 0
By substituting these values into the cross product formula, we get:
v x B = (0 - 3j + 2i) + (-4k - 12i) + (2k - j)
= (-10i - 4k - 4j)
Now, let's calculate the magnitude of the cross product vector:
|v x B| = sqrt((-10)^2 + (-4)^2 + (-4)^2)
= sqrt(100 + 16 + 16)
= sqrt(132)
= 11.49
Therefore, the magnitude of the magnetic force experienced by the proton is approximately 11.49 N.