Two parallel co-axial circular coils of equal radius ‘R’ and equal number of turns ‘N’, carry equal currents ‘I’ in the same direction and are separated by a distance ‘2R’. Find the magnitude and direction of the net magnetic field produced at the mid-point of the line joining their centres.

Solution

Given, two parallel co-axial circular coils of equal radius 'R' and equal number of turns 'N' carry equal currents 'I' in the same direction seperated by distance '2R' .

Magnetic field at the mid-point due to loop 1 is given by,

B_{1} = \frac{μ_{0} i R^{2}}{2 {(R^{2} + R^{2})}^{3 / 2}},

acting towards right .

Magnetic field at the mid-point due to loop 2 is,

B_{2} = \frac{μ_{0} i R^{2}}{2 {(R^{2} + R^{2})}^{3 / 2}},

acting towards right.

∴ Net resultant magnetic field at the mid-point is,

B = B_{1} + B_{2}

B = \frac{μ_{0} i R^{2}}{{(2 R^{2})}^{3 / 2}}

which is acting towards right.

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

Some More Questions From Moving Charges And Magnetism Chapter

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