Edit to “Solution”

astrosander edited Dec 23, 2024, 10:49 PM

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revision #12551 parent #11404 ← older newer →

en/9.1.1.md +1 −1

			### Statement

			$9.1.1.$ A force of 15 N acts on a linear conductor of length $10$ cm, located perpendicular to a magnetic field, if the current in the conductor is $1.5$ A. Find the induction of the magnetic field.
		@@ -4,8 +4,8 @@Statement

			### Solution

		−	We find the Ampere force as $$ \vec{F} = (\vec{Il})\times \vec{B} $$ In scalar form, this expression can be rewritten as $$ F = B\cdot Il \cdot \sin\alpha~~\quad(1)~~ $$ since the field is perpendicular to the current element, then $$ \alpha =90^{\circ};\quad \sin\alpha =1 $$ From here, we rewrite the expression $(1)$ as $$ F = B \cdot Il \Rightarrow \boxed{B = \frac{F}{Il} = 100\text{ T}} $$
		+	We find the Ampere force as $$ \vec{F} = (\vec{Il})\times \vec{B} $$ In scalar form, this expression can be rewritten as $$ F = B\cdot Il \cdot \sin\alpha\tag{1} $$ since the field is perpendicular to the current element, then $$ \alpha =90^{\circ};\quad \sin\alpha =1 $$ From here, we rewrite the expression $(1)$ as $$ F = B \cdot Il \Rightarrow \boxed{B = \frac{F}{Il} = 100\text{ T}} $$

			#### Answer

			$$B = \frac{F}{Il} = 100\text{ T}$$

	### Statement		### Statement

	$9.1.1.$ A force of 15 N acts on a linear conductor of length $10$ cm, located perpendicular to a magnetic field, if the current in the conductor is $1.5$ A. Find the induction of the magnetic field.		$9.1.1.$ A force of 15 N acts on a linear conductor of length $10$ cm, located perpendicular to a magnetic field, if the current in the conductor is $1.5$ A. Find the induction of the magnetic field.
	@@ -4,8 +4,8 @@Statement

	### Solution		### Solution

	We find the Ampere force as $$ \vec{F} = (\vec{Il})\times \vec{B} $$ In scalar form, this expression can be rewritten as $$ F = B\cdot Il \cdot \sin\alpha~~\quad(1)~~ $$ since the field is perpendicular to the current element, then $$ \alpha =90^{\circ};\quad \sin\alpha =1 $$ From here, we rewrite the expression $(1)$ as $$ F = B \cdot Il \Rightarrow \boxed{B = \frac{F}{Il} = 100\text{ T}} $$		We find the Ampere force as $$ \vec{F} = (\vec{Il})\times \vec{B} $$ In scalar form, this expression can be rewritten as $$ F = B\cdot Il \cdot \sin\alpha\tag{1} $$ since the field is perpendicular to the current element, then $$ \alpha =90^{\circ};\quad \sin\alpha =1 $$ From here, we rewrite the expression $(1)$ as $$ F = B \cdot Il \Rightarrow \boxed{B = \frac{F}{Il} = 100\text{ T}} $$

	#### Answer		#### Answer

	$$B = \frac{F}{Il} = 100\text{ T}$$		$$B = \frac{F}{Il} = 100\text{ T}$$