First slide

Force on a current carrying wire placed in a magnetic field

Question

What is the net force on the square coil?

Moderate

A

$25 \times 10^{- 7} N moving towards wire$
B

$25 \times 10^{- 7} N moving away from wire$
C

$35 \times 10^{- 7} N moving towards wire$
D

$35 \times 10^{- 7} N moving away from wire$

Solution

Force on side BC and AD are equal but opposite, so their resultant will be zero.
$But F_{AB} = 10^{- 7} \times \frac{2 \times 2 \times 1}{2 \times 10^{- 2}} \times 15 \times 10^{- 2} = 3 \times 10^{- 6} N$
$\begin{array}{l} and F_{CD} = 10^{- 7} \times \frac{2 \times 2 \times 1}{12 \times 10^{- 2}} \times 15 \times 10^{- 2} \\ = 0 .5 \times 10^{- 6} N \\ \Rightarrow F_{net} = F_{AB} - F_{CD} = 2 .5 \times 10^{- 6} N \\ = 25 \times 10^{- 7} N towards the wire \end{array}$

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