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A force of 1.5 × 102 N is exerted on a charge of 1.4 × 10–7 C that is traveling at an angle of 75° to a magnetic field.

If the charge is moving at 1.3 × 106 m/s, what is the magnetic field strength?

A) 8.2 × 102 T
B) 8.5 × 102 T
C) 3.2 × 103 T
D) 6.4 × 1010 T

Respuesta :

Ammimochi

The magnetic field strength would be; B) 8.5 × 102 T

Answer:

B) [tex]B = 8.5 \times 10^2 T[/tex]

Explanation:

As we know that force on a moving charge is given by

[tex]\vec F = q(\vec v \times \vec B)[/tex]

here we know that the angle between velocity and magnetic field is 75 degree

so force is given as

[tex]F = qvBsin75[/tex]

here we will have

[tex]F = 1.5 \times 10^2 N[/tex]

[tex]q = 1.4 \times 10^{-7} C[/tex]

[tex]v = 1.3 \times 10^6 m/s[/tex]

now from above formula we will have

[tex]1.5 \times 10^2 = (1.4 \times 10^{-7})(1.3 \times 10^6)Bsin75[/tex]

now from above equation

[tex]B = 8.5 \times 10^2 T[/tex]

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