Question

An alpha particle (charge +2e) and an electron move in opposite directions from the same point, each with the speed of 2.80×10^5 m/s . Find the total magnetic field these charges produce at point P, which is 8.25 nm from each charge. μ0=4πx10^-7 H/m, 1nm= 10^-9 m.

Answer 1

Answer:

Explanation:

Magnetic field due to moving charge =

$B=\frac{\mu_0}{4\pi} \times \frac{qv}{r^2}$

q is charge moving with velocity v and r is distance from point at which field is calculated .

For alpha particle

$B_1=\frac{\mu_0}{4\pi} \times \frac{2\times 1.6\times10^{-19}\times 2.8\times 10^5}{(8.25\times 10^{-9})^2}$

= 0.1316 x 10⁻³ T

For electron

$B_2=\frac{\mu_0}{4\pi} \times \frac{ 1.6\times10^{-19}\times 2.8\times 10^5}{(8.25\times 10^{-9})^2}$

= .0658 x 10⁻³ T .

Both these magnetic field will be same in direction because direction of equivalent current is same for both the particles .

Hence

Total magnetic field

= B₁ + B₂ = .1974 x 10⁻³

= 1.974 x 10⁻⁴ T .

Answer 2

The total magnetic field produced by these charges at the given point is 1.98 x 10⁻⁴ T.

The given parameters;

• charge of the alpha-particle, q = 2e
• speed of the charges, v = 2.8 x 10⁵ m/s
• separation distance, r = 8.25 nm

The magnetic field produced by each charge is calculated using Biot-Savart law;

$B = \frac{\mu _o q}{4\pi } \times \frac{v}{r^2} \\\\B_1 = \frac{(4\pi \times 10^{-7} ) \times (1.602 \times 10^{-19})}{4\pi } \times \frac{2.8 \times 10^5}{(8.25 \times 10^{-9} )^2}\\\\B_1 = 6.6 \times 10^{-5} \ T$

$B_2 = \frac{(4\pi \times 10^{-7} ) \times (2\times 1.602 \times 10^{-19})}{4\pi } \times \frac{2.8 \times 10^5}{(8.25 \times 10^{-9} )^2}\\\\B_2 = 1.32 \times 10^{-4} \ T$

The total magnetic field produced by these charges at the given point P;

$B_T = B_1 + B_2\\\\B_T = 6.6\times 10^{-5} \ + \ 1.32 \times 10^{-4}\\\\B_T = 1.98 \times 10^{-4} \ T$

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