Question
Figure shows a coil of radius 2 cm concentric with a coil of radius 7 cm Each coil has 1000 turns with a current of 5 A. In larger coil, then the current needed in the smaller coil to give the total magnetic field at the centre equal to 2 mT is
- 1.49A
- 1.84A
- 2.88A
- 3.4A
The correct answer is: 1.49A
Related Questions to study
In the given figure the magnetic induction at the point O is
In the given figure the magnetic induction at the point O is
A current IA is flowing in the sides of equilateral triangle of side 
m The magnetic induction at centroid of the triangle is
A current IA is flowing in the sides of equilateral triangle of side m The magnetic induction at centroid of the triangle is
The magnetic field at the centre of circular loop in the circuit shown below is
The magnetic field at the centre of circular loop in the circuit shown below is
Is this quadrilateral a trapezoid?
Is this quadrilateral a trapezoid?
A thin 50 cm long metal bar with mass 750 g rests on, but is not attached to two metallic supports in a uniform 0.45T magnetic field as shown in Fig .A battery and a 25
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A thin 50 cm long metal bar with mass 750 g rests on, but is not attached to two metallic supports in a uniform 0.45T magnetic field as shown in Fig .A battery and a 25resistor in series are connceted to the supports. The largest voltage the battery can have without breaking the circuit at the supports (units are in ”V”) is
What is the area of the trapezoid?
What is the area of the trapezoid?
The radius of a circle is 7 meters. What is area of the circle?
The radius of a circle is 7 meters. What is area of the circle?
Multiply: 9 × 4 =
Multiply: 9 × 4 =
Solve the quadratic equation below using the Quadratic Formula
Solve the quadratic equation below using the Quadratic Formula
In the figure, 
If AC = 75 cm AB = 100 cm and BD = 1.25 cm then AD = 
we can use the property that ratio of sides remains same in similar triangles.
In the figure, If AC = 75 cm AB = 100 cm and BD = 1.25 cm then AD =
we can use the property that ratio of sides remains same in similar triangles.
we can use the property that ratio of sides remains same in similar triangles.
we can use the property that ratio of sides remains same in similar triangles.
In the figure
and area
The value of
In the figureand areaThe value of
In the figure, 
, AB = 9 cm, AD = 7 cm, CD = 8 cm and CE = 10 cm Then DE = ?
if 2 angles are same in a triangle, then the triangles are similar. we can use the property that ratio of sides remains same in similar triangles.
In the figure, , AB = 9 cm, AD = 7 cm, CD = 8 cm and CE = 10 cm Then DE = ?
if 2 angles are same in a triangle, then the triangles are similar. we can use the property that ratio of sides remains same in similar triangles.
In the figure
PR = 6 cm then QR = ? cm
we can use the property that ratio of sides remains same in similar triangles
In the figure PR = 6 cm then QR = ? cm
we can use the property that ratio of sides remains same in similar triangles
In the figure
OB = 2x + 1, OC = 5x – 3, OD = 6x – 5 then AC = ? units.
solving the quadratic equations by the factorization method is used. in this method, the linear term is broken down into 2 terms so that we can take out the common factors from the terms and convert the equation into product form.
In the figure OB = 2x + 1, OC = 5x – 3, OD = 6x – 5 then AC = ? units.
solving the quadratic equations by the factorization method is used. in this method, the linear term is broken down into 2 terms so that we can take out the common factors from the terms and convert the equation into product form.
