Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.54 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2230 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.25 V/m, (b) in the negative z direction and has a magnitude of 5.25 V/m, and (c) in the positive x direction and has a magnitude of 5.25 V/m

Answer 1

Answer:

(a). The magnitude of the net force is $(2.1\times10^{-18}\ N)k$

(b). The magnitude of the net force is $(4.23\times10^{-19}\ N)k$

(c). The magnitude of the net force is $(8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k$

Explanation:

Given that,

Magnetic field $B=-3.54\times10^{-3}i\ T$

Velocity = 2230j m/s

We know that,

The net force acting on the proton is equal to the sum of electric and magnetic force.

$F=F_{e}+F_{B}$

(a). If the electric field is in the positive z direction and has a magnitude of 5.25 V/m,

We need to calculate the magnitude of the net force acting on the proton

Using formula of net force

$F_{net}=e(E+v\times B)$

Put the value into the formula

$F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(j\times i))$

$F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(-k))$

$F_{net}=(2.1\times10^{-18}\ N)k$

(b). If the electric field is in the negative z direction and has a magnitude of 5.25 V/m,

We need to calculate the magnitude of the net force acting on the proton

Using formula of net force

$F_{net}=e(E+v\times B)$

Put the value into the formula

$F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(j\times i))$

$F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(-k))$

$F_{net}=(4.23\times10^{-19}\ N)k$

(c). If the electric field is in the positive x direction and has a magnitude of 5.25 V/m

We need to calculate the magnitude of the net force acting on the proton

Using formula of net force

$F_{net}=e(E+v\times B)$

Put the value into the formula

$F_{net}=1.6\times10^{-19}(5.25i+2230\times-3.54\times10^{-3}(j\times i))$

$F_{net}=(8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k$

Hence, (a). The magnitude of the net force is $(2.1\times10^{-18}\ N)k$

(b). The magnitude of the net force is $(4.23\times10^{-19}\ N)k$

(c). The magnitude of the net force is $(8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k$