Question

4. Find the direction of the force on an electron moving as shown through the magnetic field pictured in the figure below. A. left B. right C. into the page D. out of the page 5. A particle of charge 3.0 C moves through a magnetic field of 1.5 T at 7.8 x 10 m/s and at a 30° angle from the magnetic field. What is the magnitude of the force? A. 0.18 N B. 0.24 N C. 0.32 N D. 0.38 N 6. What is the force between two 1.5 m parallel wire that are 10 em apart each carrying a current of 9 A in the same direction? A. 9.8 x 10-'N B.2.1 x 10-'N C.6.1 x 10-'N D.2.4 x 10-'N

Answer 1

4:- Using right hand thumb rule and since the charge ion the electron is negative the direction of the force ion the electron comes out to be towards Right.

5:- F = qvBsinθ

F = 3 * 10^-6 * 7.8 * 10⁴ * 1.5 * sin 30⁰

F = 17.55 * 10^-2 = 0.1755

Hence rounding off to two digits the ans is A. 0.18N

6:- Force between to parallel wires is given by

F = μ₀I₁I₂L/2πr

F = 2*10^-7 9 * 9 * 1.5 /0.01

F = 2.43 * 10 ^-4 N

Ans D. 2.4 * 10 ^-4 N