Q81 / 90 · JEE Main 2019 (12 Apr Shift 2) · Physics

Find the magnetic field at point $P$ due to a straight line segment $A B$ of length $6 c m$ carrying a current of $5 A .$ (See figure) $(μ_{0} = 4 π \times 10^{- 7} {NA}^{- 2})$

Correct answer
A
$1.5 \times 10^{- 5} T$
B
$2.0 \times 10^{- 5} T$
C
$3.0 \times 10^{- 5} T$
D
$2.5 \times 10^{- 5} T$

Solution

12.04.2019 Shift 2 Synergy JEE Mains D43

$θ = 37^{0}$
$r = 4 c m$
$B = \frac{μ_{0} i}{4 π r} [s i n 37 ° + s i n 37 °]$
$= \frac{4 π \times 10^{- 7} (5)}{4 π \times (4 \times 10^{- 2})} (\frac{3}{5} + \frac{3}{5})$
$= \frac{3}{2} \times 10^{- 5} = 1.5 \times 10^{- 5} T$

Hence, the correct answer is option: 'A'.

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Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A. (See figure) μ0=4π×10-7 NA-2

Solution

Relevant Questions

Find the magnetic field at point $P$ due to a straight line segment $A B$ of length $6 c m$ carrying a current of $5 A .$ (See figure) $(μ_{0} = 4 π \times 10^{- 7} {NA}^{- 2})$