First slide

Force on a current carrying wire placed in a magnetic field

Question

Two long conductors 10 cm apart carry currents in the ratio 1 :2 in the same direction. The magnetic field midway between them is $2 \times 10^{- 3} T$ The force on unit length of any one conductor will be

Moderate

A

$2 π \times 10^{- 3} N$
B

$2 \times 10^{- 3} N$
C

$π N$
D

1 N

Solution

Magnetic field at their mid-point,
$B = \frac{μ_{0}}{2 π (\frac{d}{2})} [2 i - i] = \frac{μ_{0} i}{πd}$
$Given \frac{μ_{0} i}{πd} = 2 \times 10^{- 3} T$
Substituting d = 0.1 m, we get $i = \frac{10^{3}}{2} A$
Magnetic field at the location of second conductor due to first conductor is, $B = \frac{μ_{0} i}{2 πd}$
Force on unit length of second conductor
$= B \times 2 i = \frac{μ_{0} i^{2}}{πd} = (\frac{μ_{0} i}{πd}) \times i = 1 N$

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