A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Solution

Given, current, I= 90 A
distance ,r = 1.5 m
Thus,
the magnitude of magnetic field is given by Ampere's circuital law

$B = \frac{μ_{0} I}{2 πr} = \frac{4 π \times 10^{- 7} \times 90}{2 π \times 1.5} = 1.2 \times 10^{- 5} T$

Applying the right-hand thumb rule, we find that the magnetic field at the observation point is directed towards south.

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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Moving Charges And Magnetism

A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Some More Questions From Moving Charges And Magnetism Chapter

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