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Standard XII

Physics

Motional EMF in an Open Conductor

The force act...

Question

The force acting on a pole of strength of 20 Am is 8x10 $^{- 4}$ N, when placed in a magnetic field. The magnetic induction field strength is :

\times

^{- 6}

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0.4

\times

^{4}

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4 G

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\times

^{- 5}

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Solution

The correct option is C 4 $\times$ 10 $^{- 5}$ T
$m = 20 A m$
$F = 8 \times 10^{- 4} N$
$F = B m$
$8 \times 10^{- 4} = 20 \times B$
$B = 4 \times 10^{- 5} T$

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The force acting on a pole of strength of 20 Am is 8x10−4 N, when placed in a magnetic field. The magnetic induction field strength is :

The correct option is C 4×10−5 Tm=20AmF=8×10−4NF=Bm8×10−4=20×BB=4×10−5T

The force acting on a pole of strength of 20 Am is 8x10 $^{- 4}$ N, when placed in a magnetic field. The magnetic induction field strength is :

The correct option is C 4 $\times$ 10 $^{- 5}$ T
$m = 20 A m$
$F = 8 \times 10^{- 4} N$
$F = B m$
$8 \times 10^{- 4} = 20 \times B$
$B = 4 \times 10^{- 5} T$