A straight horizontal conducting rod of 30 cm and mass 90 grams is suspended by two vertical wires at its ends. A current of 8A is setup in rod through wires. What should be the magnetic field (setup normal to conductor) so that tension in wires become zero.

\frac{3}{8}

Tesla

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\frac{1}{4}

Tesla

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\frac{8}{3}

Tesla

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4 Telsa

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Solution

The correct option is A $\frac{3}{8}$ Tesla

For eqbm
$2 T + b i l - m g = 0$
$2 T = m g - B i l$
if $T = 0, then, m g - B i l = 0$
$B = \frac{m g}{i l} = \frac{90 \times 10^{- 3} \times 10}{8 \times 30 \times 10^{- 2}} = \frac{9 \times 10^{- 1}}{8 \times 3 \times 10^{- 1}} = \frac{3}{8} T e s l a$

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Similar questions

Q. A straight horizontal conducting rod of 30 cm and mass 90 grams is suspended by two vertical wires at its ends. A current of 8A is setup in rod through wires. What should be the magnetic field (setup normal to conductor) so that tension in wires become zero.

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.

A straight horizontal conducting rod of length 0.5 m and mass 50 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. What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero? Ignore the mass of the wires and take g = 10 $m s^{- 2}$ .

Q. A straight horizontal conducting rod of length

0.5 m

and mass

50 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. Then the magnetic field set up normal to the conductor in order that the tension in the wires is zero is (ignore the mass of the wire and take

g = 10 m s^{- 2}

) :

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A straight horizontal conducting rod of 30 cm and mass 90 grams is suspended by two vertical wires at its ends. A current of 8A is setup in rod through wires. What should be the magnetic field (setup normal to conductor) so that tension in wires become zero.

The correct option is A 38 Tesla For eqbm 2T+bil−mg=0 2T=mg−Bil if T=0, then, mg−Bil=0 B=mgil=90×10−3×108×30×10−2=9×10−18×3×10−1=38Tesla

The correct option is A $\frac{3}{8}$ Tesla

For eqbm
$2 T + b i l - m g = 0$
$2 T = m g - B i l$
if $T = 0, then, m g - B i l = 0$
$B = \frac{m g}{i l} = \frac{90 \times 10^{- 3} \times 10}{8 \times 30 \times 10^{- 2}} = \frac{9 \times 10^{- 1}}{8 \times 3 \times 10^{- 1}} = \frac{3}{8} T e s l a$