Question

An electron that is moving through a uniform magnetic field has
a velocity v = (52 *km/s*)**i** + 40
*km/s*)**j** when it experiences a force F =
(-3.600×10^{-15} *N*)**i** +
(4.680×10^{-15} *N*)**j**due to the
magnetic field. If Bx = 0, calculate the magnitude of the magnetic
field B. (Tesla = T). What is By? What is Bz?

Answer #1

^{3} m/s

F =( -3.6 i + 4.68 j )*10^{-15} N

We know that

Magnetic force on the charge is given by

F = q [V x B]

B = B_{X} i + B_{Y} j + B_{z} k

Where B_{x} = 0 , therefore

B = 0 i + B_{Y} j + B_{Z} k

where q is charge on electron = 1.6*10^{-19}

F = (1.6*10^{-19})[( 52*10^{3} i +
40*10^{3} j ) x (0 i + B_{Y} j + B_{Z} k )
]

On solving we get

( -3.6 i + 4.68 j )*10^{-15} =
1.6*10^{-19}*10^{3}[40 B_{Z} i -
52B_{Z} j + 52B_{Y} k ]

On compairing

B_{Y} = 0

B_{Z} = -0.5625 T

Now the magnitude of the B

B = 0.5625 T

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