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3 years ago

8

Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.650 µC charge and flies due west at a speed of

540 m/s over the Earth's south magnetic pole, where the 8.00 ✕ 10−5 T magnetic field points straight up. What are the magnitude (in N) and direction of the magnetic force on the plane?

1 answer:

Ganezh [65]3 years ago

4 0

Answer : $F=2808\times 10^{-11}\ N$

Explanation :

It is given that,

Charge, $q=0.650\ \mu C=0.65\times 10^{-6}\ C$

Velocity of Aircraft, $v=540\ m/s$ (in west)

Magnetic field, $B=8 \times 10^{-5}\ T$ ( in north )

Using the relation as :

$F=q(v\times B)$

Magnetic force is ,

$F=0.65\times 10^{-6}\ C\times 540\ m/s\times 8 \times 10^{-5}\ T$

$F=2808\times 10^{-11}\ N$

Using Right hand thumb rule, the direction of force is in the plane perpendicular to the velocity and the magnetic field i.e. $-\hat{k}$ .

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