16. Find the magnetic field due to a bar magnet of length 10 cm10\text{ cm} and pole strength 20 Am20\text{ Am} at a point 15 cm15\text{ cm} from it on axial line. a) 1.5 T1.5\text{ T} b) 1.5×10−2 T1.5 \times 10^{-2}\text{ T} c) 1.5×10−4 T1.5 \times 10^{-4}\text{ T} d) 1.5×10−6 T1.5 \times 10^{-6}\text{ T} 17. Find the magnetic induction at point PP in the given figure (The magnets are short). Figure description: Two parallel horizontal bar magnets are shown, one above the other. The top bar magnet has poles labeled N\text{N} on the left and S\text{S} on the right, with magnetic moment M=512 Am2M = 512\text{ Am}^2. The bottom bar magnet has poles labeled N\text{N} on the left and S\text{S} on the right, with magnetic moment M=1024 Am2M = 1024\text{ Am}^2. A vertical line connects the centers of both magnets, perpendicular to their axes (along their equatorial lines). Point PP lies on this line, at a distance of 1 m1\text{ m} below the center of the top magnet and 1 m1\text{ m} above the center of the bottom magnet. a) 1.5×10−4 T1.5 \times 10^{-4}\text{ T} b) 3×10−4 T3 \times 10^{-4}\text{ T} c) 6×10−4 T6 \times 10^{-4}\text{ T} d) None

Answer: 16. c) 1.5×10−4 T1.5 \times 10^{-4}\text{ T}, 17. a) 1.5×10−4 T1.5 \times 10^{-4}\text{ T}

Step-by-step solution

Step 1: Identify parameters for Question 16

For the bar magnet in Question 16, the total length is 2l=10 cm2l = 10\text{ cm}, which gives semi-length l=5 cm=0.05 ml = 5\text{ cm} = 0.05\text{ m}. The pole strength is m=20 A⋅mm = 20\text{ A}\cdot\text{m}, and the distance from the center of the magnet to the axial point is d=15 cm=0.15 md = 15\text{ cm} = 0.15\text{ m}. The magnetic dipole moment is M=m×2l=20×0.1=2 A⋅m2M = m \times 2l = 20 \times 0.1 = 2\text{ A}\cdot\text{m}^2.

Step 2: Calculate axial magnetic field for Question 16

Using the axial field formula for a finite bar magnet, we substitute d2−l2=0.0225−0.0025=0.02 m2d^2 - l^2 = 0.0225 - 0.0025 = 0.02\text{ m}^2. Then (d2−l2)2=(0.02)2=4×10−4 m4(d^2 - l^2)^2 = (0.02)^2 = 4 \times 10^{-4}\text{ m}^4. The numerator is 10−7×2×2×0.15=6×10−810^{-7} \times 2 \times 2 \times 0.15 = 6 \times 10^{-8}. Dividing gives Baxial=6×10−84×10−4=1.5×10−4 TB_{\text{axial}} = \frac{6 \times 10^{-8}}{4 \times 10^{-4}} = 1.5 \times 10^{-4}\text{ T}. Thus, the correct option for Question 16 is (c).

Step 3: Analyze equatorial magnetic field for Question 17

Point PP lies on the equatorial line of both short bar magnets. For a short bar magnet, the magnetic field at a distance rr on the equatorial line is directed opposite to the magnetic dipole moment M⃗\vec{M}. Since both magnets have their North poles on the left and South poles on the right, both dipole moments M⃗1\vec{M}_1 and M⃗2\vec{M}_2 point from South to North, which is towards the left. Consequently, both equatorial fields B⃗1\vec{B}_1 and B⃗2\vec{B}_2 point in the opposite direction, towards the right.

Step 4: Calculate resultant field at point P for Question 17

Since both fields B⃗1\vec{B}_1 and B⃗2\vec{B}_2 point in the same direction (to the right), the net magnetic induction is the scalar sum: Bnet=B1+B2B_{\text{net}} = B_1 + B_2. Substituting the values M1=512 A⋅m2M_1 = 512\text{ A}\cdot\text{m}^2, M2=1024 A⋅m2M_2 = 1024\text{ A}\cdot\text{m}^2, and r=1 mr = 1\text{ m}, we get Bnet=10−7×(512+1024)=10−7×1536=1.536×10−4 T≈1.5×10−4 TB_{\text{net}} = 10^{-7} \times (512 + 1024) = 10^{-7} \times 1536 = 1.536 \times 10^{-4}\text{ T} \approx 1.5 \times 10^{-4}\text{ T}. This corresponds to option (a).

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