A particle with a charge of - 1.24 x 10"° C is moving with instantaneous velocity (4.19 X 104 m/s) î + (-3.85 X 104 m/s) j. What is the force exerted on this particle by a magnetic field (a) = B (1.40 T) i and (b) È = (1.40 T) k?

(a) F = 6.68*10¹¹⁴ N (-k)

(b) F = (6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q * v X B Formula (1)

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q = - 1.24 * 10¹¹⁰ C

v = (4.19 * 10⁴ m/s) î + (-3.85 * 10⁴m/s) j

B = (1.40 T) i

B = (1.40 T) k

Problem development

a) vXB = (4.19 * 10⁴ m/s) î + (-3.85 * 10⁴m/s) j X (1.40 T) i =

= - (-3.85*1.4) k = 5.39 * 10⁴ m/s*T (k)

1T = 1 N / C*m/s

We apply the formula (1)

F = 1.24 * 10¹¹⁰ C * 5.39 * 10⁴ m/s * N / C*m/s (-k)

F = 6.68*10¹¹⁴ N (-k)

a) vXB = (4.19 * 10⁴ m/s) î + (-3.85 * 10⁴m/s) j X (1.40 T) k =

= ( - 5.39 * 10⁴i - 5.87 * 10⁴j) m/s*T

1T = 1 N / C*m/s

We apply the formula (1)

F = 1.24 * 10¹¹⁰ C * (5.39 * 10⁴i + 5.87 * 10⁴j) m/s * N / C*m/s

F = (6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j) N