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Question

A point charge of magnitude q = 4.5 nC is moving with speed v = 3.6×107 m/s parallel to the x-axis along the line y = 3 m. Find the magnetic field at the origin produced by this charge when the charge is at the point x = -4 m, y = 3 m, as shown in the figure.


A
3.89×1010^k T
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B
3.89×1010^k T
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C
3.89×1010^i T
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D
3.89×1010^j T
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Solution

The correct option is A 3.89×1010^k T
The magnetic field is given by B=μ04π qv×^rr2,with v=v^i
r=(4^i3^j)r=42+32=5 m

Unit vector in the direction of ^r=rr=4^i3^j5=0.8^i0.6^j
v×^r=(v^i)×(0.8^i0.6^j)=0.6v^k

B=μ04π qv×^rr2=μ04π q(0.6v^k)r2=(107)(4.5×109)(0.6)(3.6×107)52^k=3.89×1010T ^k

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