Question

A solenoid with 200 loops is 50 cm long. It has a current 1.2 A. What is the magnetic field?

An electron moving along the x-axis with a velocity of  v⇀ = 2.5 x 10⁶ î m/s enters a region of magnetic field: B⇀ = 2.26 ĵ  T (teslas).  What is the force (magnitude in newtons + direction) on the electron?

Answer 1

Part A.

Magnetic field inside a solenoid is given by:

B = *n*i

n = number of turns per unit length = N/L

B = *N*i/L

Using given values:

B = 4*pi*10^-7*200*1.2/0.50

B = 0.603*10^-3 T = 0.603 mT

Part B.

Magnetic force on a moving charge particle in magnetic field is given by:

F = q*VxB

q = charge on electron = -e = -1.6*10^-19 C

VxB = (2.5*10^6 i)x(2.26 j) = 2.5*2.26*10^6*(ixj)

VxB = (5.65*10^6 k)

So,

F = -1.6*10^-19*5.65*10^6 k

F = (-9.04*10^-13 k) N

|F| = magnitude of force = 9.04*10^-13 N

Direction = -k = Into the page