A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before?
(Ignore the mass of the wires.) g = 9.8 m s^–2.

Solution

Given,
Length of the rod, l= 0.45 m
mass of the suspension, m = 60 g = 60 x 10^–3 kg
Current carried by the conductor, I = 5.0 A

(a) Force needed to balance the weight of the rod,

$F = mg = 0.06 kg \times 9.8 = 0.588 N$

Using the formula,

$F = BI l$

$B = \frac{F}{I l} = \frac{0.588}{5.0 \times 0.45} = 0.26 T$

If the direction of current is from right to left then the direction of magnetic field is horizontal and normal to the conductor and the force due to magnetic field will be upwards (Fleming's left hand rule).

(b) If the direction of current is reversed then ‘BIl’ and ‘mg’ will act vertically downwards.
Therefore,
Total tension in the wires = BIl + mg
= 0.588 + 0.588
= 1.176 N.

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

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