Physics Question for those geniusess
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Physics Question for those geniusess

[From: ] [author: ] [Date: 12-05-03] [Hit: ]
For Second part F=q*(VxB),which modifies to F=q*V*B*sin(Angle).The angle comes out to be 22.5 degrees.......
Please help me with these two please..

1) A wire 2.6 m in length carries a current of 8.4 A in a region where a uniform magnetic field has a magnitude of 0.65 T. Calculate the magnitude of the magnetic force on the wire if the angle between the magnetic field and the current is 17◦
Answer in units of N

2) A proton moving at 7.6 × 10 ^ 6 m/s through a magnetic field of 3 T experiences a magneticforce of magnitude 1.4 × 10 ^ −12 N.
What is the angle between the proton’s velocity and the field? The charge on a
proton is 1.60218 × 10 ^−19 C and its mass is 1.67262 × 10 ^ −27 kg.
Answer in units of
◦ degree

Thank u so much guys thankk u

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For first part use the equation F=i*LxB, where B is the magnetic field, i is the current, L is the length.
As it's a cross product and angle is also given hence the equation modifies to F=i*(L*B)*sin(Angle).
Which comes out to be 4.15 N
For Second part F=q*(VxB), which modifies to F=q*V*B*sin(Angle).
The angle comes out to be 22.5 degrees.
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