JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 26)
Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5A.
(See figure) ($$\mu $$0 = 4$$\pi $$ × 10–7 N-A–2)
_12th_April_Evening_Slot_en_26_1.png)
1.5 × 10–5
T
3.0 × 10–5
T
2.0 × 10–5
T
2.5 × 10–5
T
Explanation
_12th_April_Evening_Slot_en_26_2.png)
B = $${{{\mu _0}I} \over {4\pi r}}\left( {\sin {\theta _1} + \sin {\theta _2}} \right)$$
As $$\theta $$1 = $$\theta $$2 = $$\theta $$
B = $${{{\mu _0}I} \over {4\pi r}}\left( {2\sin \theta } \right)$$
= $${{{{10}^{ - 7}} \times 5} \over {4 \times {{10}^{ - 2}}}} \times 2 \times {3 \over 5}$$
= 1.5 $$ \times $$ 10-5 T
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