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Magnetic field due to current element

Question

An elevator carrying a charge of 0.5C is moving down with a velocity of 5 x 103 ms-1. The elevator is 4 m  from the bottom and 3 m horizontally away from P as shown in figure. What magnetic field (in μT) does it produce at point P?

Moderate
Solution

According to Biot-Savart law,

B=μ04πqvr2sinθ=10-7×0.5×5×1035235=6μT



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A charge particle A of charge q = 2C has velocity V = 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of instantaneous magnetic field at point B (r = 2m) due to this moving charge is  0.5×n  μT. Find n.


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