Question

a) A wire 40 cm long conducts a current of 2A perpendicular to the direction of a magnetic field B=0.25T. Find the magnetic force over the wire.

b) An alpha particle (+2e) its projected into a magnetic field of 0.14T with a velocity of 3.9x10⁶m/s. Find the magnetic force over the charge at the instant in which the direction of its velocity forms an angle of 55° with respect to the magnetic flux.

Answer 1

(a) we have the parameters

Length of wire, $l$ = 40cm =0.4m

Current, $I$ = 2A

Magnetic field, $B$ = 0.25T

We also have that, the current is perpendicular to the direction of the magnetic field

Therefore, $\theta$= $90°$

We use the formula

$F=BIlsinθ$

$=0.25*2*0.4*sin90°$

$=0.25*2*0.4*1$(since $Sin90°=1$)

$=0.25*2*0.4$

$=0.25*0.8$

$=0.2$$N$

(b) We have the parameters;

Charge, q = +2e

Magnetic field, B =0.14T

Velocity, v = $3.9*10^{-6}$ m/s

Angle, $\theta=55°$

Now, we use the magnetic force formula

$F=qvB*sin\theta$

$F=2*3.9*10^{-6}*0.14*sin55°$

$F=2*3.9*10^{-6}*0.14*0. 8192$

$F=0.229376*3.9*10^{-6}$

$F=8.9457*10^{-7}N$