A charged particle A of charge q = 2 C has velocity v = 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of magnetic field at point B due to this moving charge is (r = 2 m).the agle between them is 30.
2.5μT
5.0μT
2.0μT
None
A
2.5μT
B
5.0μT
C
2.0μT
D
None
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Solution
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μ04πq→v×→rr3=μ04πqvsinθr2
=μ04π2×100×sin3004
=10−7×25
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