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The force required to separate two surfaces with a contact area measuring 3 cm × 4 cm, when flux density normal to the surface is 0.6 tesla, will be
- A735.4 N
- B171.8 N
- C275.8 N
- D147.6 N
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CCommunity
Given that, Area = 3 × 10-2 × 4 × 10-2 = 12 × 10-4
Flux density (B) = 0.6 T
The force is given by, $(F = \frac{{{B^2}A}}{{2{\mu _0}}})$
$(= \frac{{{{0.6}^2} \times 12 \times {{10}^{ - 4}}}}{{2 \times 4\pi \times {{10}^{ - 7}}}} = 171.8\;N)$
Flux density (B) = 0.6 T
The force is given by, $(F = \frac{{{B^2}A}}{{2{\mu _0}}})$
$(= \frac{{{{0.6}^2} \times 12 \times {{10}^{ - 4}}}}{{2 \times 4\pi \times {{10}^{ - 7}}}} = 171.8\;N)$