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Home » A straight horizontal conducting rod of length and mass is suspended by two vertical wires at its ends. A current of 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 the same as before? (Ignore the mass of the wires.) .
A straight horizontal conducting rod of length 0.45 \mathrm{~m} and mass 60 \mathrm{~g} is suspended by two vertical wires at its ends. A current of 5.0 \mathrm{~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 the same as before? (Ignore the mass of the wires.) \mathbf{g}=9.8 \mathrm{~m} \mathrm{~s}^{-2}.
CBSE | Class 12 | Moving Charges and Magnetism | NCERT | Physics
A straight horizontal conducting rod of length 0.45 \mathrm{~m} and mass 60 \mathrm{~g} is suspended by two vertical wires at its ends. A current of 5.0 \mathrm{~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 the same as before? (Ignore the mass of the wires.) \mathbf{g}=9.8 \mathrm{~m} \mathrm{~s}^{-2}.

Length of the rod, I =0.45 \mathrm{~m}

Mass suspended, m=60 \mathrm{~g}=60 \times 10^{-3} \mathrm{Kg}

Current, \mid=5 \mathrm{~A}

(a) Tension in the wire is zero if the magnetic field’s force on the current-carrying wire is equal to and opposite to the weight on the wire.

B |=m g

\mathrm{B}=\mathrm{mg} / \mathrm{ll}

=\frac{60 \times 10^{-3} \times 9.8}{5 \times 0.45}=0.26133 \mathrm{~T}

The magnetic field that has been built up normal to the conductor is 0.26133 \mathrm{~T}

(b) When the current direction is reversed, BII and mg will act vertically downwards, increasing the effective tension in the wire.

\mathrm{T}=\mathrm{B} |+\mathrm{mg}

=(0.26133 \times 5.0 \times 0.45)+\left(60 \times 10^{-3} \times 9.8\right)=0.587+(0.588)=1.176 \mathrm{~N}

Total tension in the wire =1.176 \mathrm{~N}

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