Answer:
The magnitude of magnetic field is 0.312 T and direction of the magnetic field is out of the page.
Explanation:
Given:
Velocity of proton [tex]v = 1.6 \times 10^{6}[/tex] [tex]\frac{m}{s}[/tex]
Magnetic force [tex]F = 8 \times 10^{-14}[/tex] N
Charge of proton [tex]q = 1.6 \times 10^{-19}[/tex]C
The magnetic force experience by proton in magnetic field is given by,
[tex]F =qvB \sin \theta[/tex]
Here magnetic force experienced by proton is maximum so we take [tex]\sin \theta = 1[/tex]
[tex]F = qvB[/tex]
[tex]B = \frac{F}{qv}[/tex]
[tex]B = \frac{8 \times 10^{-14} }{1.6 \times 10^{-19} \times 1.6 \times 10^{6} }[/tex]
[tex]B = 0.312[/tex] T
According to the left hand rule the direction of magnetic field is out of the page.
Therefore, the magnitude of magnetic field is 0.312 T and direction of the magnetic field is out of the page.
