JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Magnetic Effects of Current

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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)
(

μ 0 = 4 p \times 10 - 7 N - A - 2

)

(A) $2.0 \times 1 0^{- 5} T$
(B) $3.0 \times 1 0^{- 5} T$
(C) $2.5 \times 1 0^{- 5} T$
(D) $1.5 \times 1 0^{- 5} T$ ✓ Correct

Answer: (D)

Solution

For a finite straight wire the field at

P

B = \frac{μ _{0} I}{4 π d} (sin θ_{1} + sin θ_{2})

. From the geometry the perpendicular distance is

d = 4

cm and each end subtends an angle with

sin θ = \frac{3}{5}

, read from the

3

–

4

–

5

triangle. So

B = \frac{1 0 ^{- 7} \times 5}{0.04} (\frac{3}{5} + \frac{3}{5}) = \frac{5 \times 1 0 ^{- 7}}{0.04} \times \frac{6}{5} = 1.5 \times 1 0^{- 5}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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