JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 10)
A solenoid of 1000 turns per metre has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5A. The magnetic flux density produced by the solenoid is : (permeability of free space = 4$$\pi$$ $$\times$$ 10$$-$$7 H/m)
$$\pi$$T
2 $$\times$$ 10$$-$$3$$\pi$$ T
10$$-$$4$$\pi$$ T
$${\pi \over 5}$$ T
Explanation
B = $$\mu$$ n i
B = $$\mu$$r $$\mu$$0 n i
B = 500 $$\times$$ 4$$\pi$$ $$\times$$ 10$$-$$7 $$\times$$ 103 $$\times$$ 5
B = $$\pi$$ $$\times$$ 10$$-$$3 $$\times$$ 103
B = $$\pi$$ T
B = $$\mu$$r $$\mu$$0 n i
B = 500 $$\times$$ 4$$\pi$$ $$\times$$ 10$$-$$7 $$\times$$ 103 $$\times$$ 5
B = $$\pi$$ $$\times$$ 10$$-$$3 $$\times$$ 103
B = $$\pi$$ T
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