Question Description
The magnetic flux density \( B \) at a distance \( d \) from a long straight current-carrying conductor is given by:\[B = \frac{\mu_0 I}{2 \pi d}\]where:- \( \mu_0 \) is the permeability of free space,- \( I \) is the current in the conductor,- \( d \) is the distance from the conductor.If the distance is halved to \( \frac{d}{2} \), the new magnetic flux density \( B' \) becomes:\[B' = \frac{\mu_0 I}{2 \pi \left(\frac{d}{2}\right)} = \frac{\mu_0 I}{\pi d} = 2 \times \frac{\mu_0 I}{2 \pi d} = 2B\]Thus, the value of \( a \) is \( \boxed{2} \).? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about The magnetic flux density \( B \) at a distance \( d \) from a long straight current-carrying conductor is given by:\[B = \frac{\mu_0 I}{2 \pi d}\]where:- \( \mu_0 \) is the permeability of free space,- \( I \) is the current in the conductor,- \( d \) is the distance from the conductor.If the distance is halved to \( \frac{d}{2} \), the new magnetic flux density \( B' \) becomes:\[B' = \frac{\mu_0 I}{2 \pi \left(\frac{d}{2}\right)} = \frac{\mu_0 I}{\pi d} = 2 \times \frac{\mu_0 I}{2 \pi d} = 2B\]Thus, the value of \( a \) is \( \boxed{2} \).? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The magnetic flux density \( B \) at a distance \( d \) from a long straight current-carrying conductor is given by:\[B = \frac{\mu_0 I}{2 \pi d}\]where:- \( \mu_0 \) is the permeability of free space,- \( I \) is the current in the conductor,- \( d \) is the distance from the conductor.If the distance is halved to \( \frac{d}{2} \), the new magnetic flux density \( B' \) becomes:\[B' = \frac{\mu_0 I}{2 \pi \left(\frac{d}{2}\right)} = \frac{\mu_0 I}{\pi d} = 2 \times \frac{\mu_0 I}{2 \pi d} = 2B\]Thus, the value of \( a \) is \( \boxed{2} \).?.

Solutions for The magnetic flux density \( B \) at a distance \( d \) from a long straight current-carrying conductor is given by:\[B = \frac{\mu_0 I}{2 \pi d}\]where:- \( \mu_0 \) is the permeability of free space,- \( I \) is the current in the conductor,- \( d \) is the distance from the conductor.If the distance is halved to \( \frac{d}{2} \), the new magnetic flux density \( B' \) becomes:\[B' = \frac{\mu_0 I}{2 \pi \left(\frac{d}{2}\right)} = \frac{\mu_0 I}{\pi d} = 2 \times \frac{\mu_0 I}{2 \pi d} = 2B\]Thus, the value of \( a \) is \( \boxed{2} \).? in English & in Hindi are available as part of our courses for Physics. Download more important topics, notes, lectures and mock test series for Physics Exam by signing up for free.

Here you can find the meaning of The magnetic flux density \( B \) at a distance \( d \) from a long straight current-carrying conductor is given by:\[B = \frac{\mu_0 I}{2 \pi d}\]where:- \( \mu_0 \) is the permeability of free space,- \( I \) is the current in the conductor,- \( d \) is the distance from the conductor.If the distance is halved to \( \frac{d}{2} \), the new magnetic flux density \( B' \) becomes:\[B' = \frac{\mu_0 I}{2 \pi \left(\frac{d}{2}\right)} = \frac{\mu_0 I}{\pi d} = 2 \times \frac{\mu_0 I}{2 \pi d} = 2B\]Thus, the value of \( a \) is \( \boxed{2} \).? defined & explained in the simplest way possible. Besides giving the explanation of The magnetic flux density \( B \) at a distance \( d \) from a long straight current-carrying conductor is given by:\[B = \frac{\mu_0 I}{2 \pi d}\]where:- \( \mu_0 \) is the permeability of free space,- \( I \) is the current in the conductor,- \( d \) is the distance from the conductor.If the distance is halved to \( \frac{d}{2} \), the new magnetic flux density \( B' \) becomes:\[B' = \frac{\mu_0 I}{2 \pi \left(\frac{d}{2}\right)} = \frac{\mu_0 I}{\pi d} = 2 \times \frac{\mu_0 I}{2 \pi d} = 2B\]Thus, the value of \( a \) is \( \boxed{2} \).?, a detailed solution for The magnetic flux density \( B \) at a distance \( d \) from a long straight current-carrying conductor is given by:\[B = \frac{\mu_0 I}{2 \pi d}\]where:- \( \mu_0 \) is the permeability of free space,- \( I \) is the current in the conductor,- \( d \) is the distance from the conductor.If the distance is halved to \( \frac{d}{2} \), the new magnetic flux density \( B' \) becomes:\[B' = \frac{\mu_0 I}{2 \pi \left(\frac{d}{2}\right)} = \frac{\mu_0 I}{\pi d} = 2 \times \frac{\mu_0 I}{2 \pi d} = 2B\]Thus, the value of \( a \) is \( \boxed{2} \).? has been provided alongside types of The magnetic flux density \( B \) at a distance \( d \) from a long straight current-carrying conductor is given by:\[B = \frac{\mu_0 I}{2 \pi d}\]where:- \( \mu_0 \) is the permeability of free space,- \( I \) is the current in the conductor,- \( d \) is the distance from the conductor.If the distance is halved to \( \frac{d}{2} \), the new magnetic flux density \( B' \) becomes:\[B' = \frac{\mu_0 I}{2 \pi \left(\frac{d}{2}\right)} = \frac{\mu_0 I}{\pi d} = 2 \times \frac{\mu_0 I}{2 \pi d} = 2B\]Thus, the value of \( a \) is \( \boxed{2} \).? theory, EduRev gives you an ample number of questions to practice The magnetic flux density \( B \) at a distance \( d \) from a long straight current-carrying conductor is given by:\[B = \frac{\mu_0 I}{2 \pi d}\]where:- \( \mu_0 \) is the permeability of free space,- \( I \) is the current in the conductor,- \( d \) is the distance from the conductor.If the distance is halved to \( \frac{d}{2} \), the new magnetic flux density \( B' \) becomes:\[B' = \frac{\mu_0 I}{2 \pi \left(\frac{d}{2}\right)} = \frac{\mu_0 I}{\pi d} = 2 \times \frac{\mu_0 I}{2 \pi d} = 2B\]Thus, the value of \( a \) is \( \boxed{2} \).? tests, examples and also practice Physics tests.