First slide

Applications involving charged particles moving in a magnetic field

Question

Three long straight and parallel wires are arranged as shown in figure. The force experienced by 10 cm length of wire ‘Q’ is

Moderate

A

$1.4 \times 10^{- 4} N$ towards the right
B

$2.6 \times 10^{- 4} N$ towards right
C

$1.4 \times 10^{- 4} N$ towards the left
D

$2.6 \times 10^{- 4} N$ toward left

Solution

$W i r e Q a n d R w i l l r e p e l a n d w i r e Q a n d P w i l l a l s o r e p e l . A s s h o w n i n f i g u r e, n e t f o r c e c a n b e g i v e n a s, F_{n e t} = F_{Q R} - F_{Q P} \Rightarrow F_{n e t} = \frac{μ_{o}}{2 π} \frac{i_{R} i_{Q} L}{d_{Q R}} - \frac{μ_{o}}{2 π} \frac{i_{Q} i_{P} L}{d_{Q P}} \Rightarrow F_{n e t} = \frac{4 π \times 10^{- 7} \times 20 \times 10 \times 10 \times 10^{- 2}}{2 π \times 2 \times 10^{- 2}} - \frac{4 π \times 10^{- 7} \times 10 \times 30 \times 10 \times 10^{- 2}}{2 π \times 10 \times 10^{- 2}} \Rightarrow F_{n e t} = \frac{4 π \times 10^{- 7} \times 10 \times 10^{- 2}}{2 π \times 10^{- 2}} [100 - 30] = 20 \times 10^{- 7} \times 70 = 1400 \times 10^{- 7} = 1.4 \times 10^{- 4} N towards right$

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